diff --git a/problems/0001.两数之和.md b/problems/0001.两数之和.md
index eea3ba7a..1af8787b 100644
--- a/problems/0001.两数之和.md
+++ b/problems/0001.两数之和.md
@@ -151,7 +151,7 @@ public int[] twoSum(int[] nums, int target) {
 ```
 
 Python：
-
+（版本一） 使用字典
 ```python
 class Solution:
     def twoSum(self, nums: List[int], target: int) -> List[int]:
@@ -163,6 +163,53 @@ class Solution:
             records[value] = index    # 遍历当前元素，并在map中寻找是否有匹配的key
         return []
 ```
+（版本二）使用集合
+```python
+class Solution:
+    def twoSum(self, nums: List[int], target: int) -> List[int]:
+        #创建一个集合来存储我们目前看到的数字
+        seen = set()             
+        for i, num in enumerate(nums):
+            complement = target - num
+            if complement in seen:
+                return [nums.index(complement), i]
+            seen.add(num)
+```
+（版本三）使用双指针
+```python
+class Solution:
+    def twoSum(self, nums: List[int], target: int) -> List[int]:
+        # 对输入列表进行排序
+        nums_sorted = sorted(nums)
+        
+        # 使用双指针
+        left = 0
+        right = len(nums_sorted) - 1
+        while left < right:
+            current_sum = nums_sorted[left] + nums_sorted[right]
+            if current_sum == target:
+                # 如果和等于目标数，则返回两个数的下标
+                left_index = nums.index(nums_sorted[left])
+                right_index = nums.index(nums_sorted[right])
+                if left_index == right_index:
+                    right_index = nums[left_index+1:].index(nums_sorted[right]) + left_index + 1
+                return [left_index, right_index]
+            elif current_sum < target:
+                # 如果总和小于目标，则将左侧指针向右移动
+                left += 1
+            else:
+                # 如果总和大于目标值，则将右指针向左移动
+                right -= 1
+```
+（版本四）暴力法
+```python
+class Solution:
+    def twoSum(self, nums: List[int], target: int) -> List[int]:
+        for i in range(len(nums)):
+            for j in range(i+1, len(nums)):
+                if nums[i] + nums[j] == target:
+                    return [i,j]
+```
 
 Go：
 
diff --git a/problems/0015.三数之和.md b/problems/0015.三数之和.md
index 26c9eaa2..9cca779b 100644
--- a/problems/0015.三数之和.md
+++ b/problems/0015.三数之和.md
@@ -298,61 +298,72 @@ class Solution {
 ```
 
 Python：
+（版本一） 双指针
 ```Python
 class Solution:
-    def threeSum(self, nums):
-        ans = []
-        n = len(nums)
+    def threeSum(self, nums: List[int]) -> List[List[int]]:
+        result = []
         nums.sort()
-	# 找出a + b + c = 0
-        # a = nums[i], b = nums[left], c = nums[right]
-        for i in range(n):
-            left = i + 1
-            right = n - 1
-	    # 排序之后如果第一个元素已经大于零，那么无论如何组合都不可能凑成三元组，直接返回结果就可以了
-            if nums[i] > 0: 
-                break
-            if i >= 1 and nums[i] == nums[i - 1]: # 去重a
+        
+        for i in range(len(nums)):
+            # 如果第一个元素已经大于0，不需要进一步检查
+            if nums[i] > 0:
+                return result
+            
+            # 跳过相同的元素以避免重复
+            if i > 0 and nums[i] == nums[i - 1]:
                 continue
-            while left < right:
-                total = nums[i] + nums[left] + nums[right]
-                if total > 0:
-                    right -= 1
-                elif total < 0:
+                
+            left = i + 1
+            right = len(nums) - 1
+            
+            while right > left:
+                sum_ = nums[i] + nums[left] + nums[right]
+                
+                if sum_ < 0:
                     left += 1
+                elif sum_ > 0:
+                    right -= 1
                 else:
-                    ans.append([nums[i], nums[left], nums[right]])
-		    # 去重逻辑应该放在找到一个三元组之后，对b 和 c去重
-                    while left != right and nums[left] == nums[left + 1]: left += 1
-                    while left != right and nums[right] == nums[right - 1]: right -= 1
-                    left += 1
+                    result.append([nums[i], nums[left], nums[right]])
+                    
+                    # 跳过相同的元素以避免重复
+                    while right > left and nums[right] == nums[right - 1]:
+                        right -= 1
+                    while right > left and nums[left] == nums[left + 1]:
+                        left += 1
+                        
                     right -= 1
-        return ans
+                    left += 1
+                    
+        return result
 ```
-Python (v3):
+（版本二） 使用字典
 
 ```python
 class Solution:
     def threeSum(self, nums: List[int]) -> List[List[int]]:
-        if len(nums) < 3: return []
-        nums, res = sorted(nums), []
-        for i in range(len(nums) - 2):
-            cur, l, r = nums[i], i + 1, len(nums) - 1
-            if res != [] and res[-1][0] == cur: continue # Drop duplicates for the first time.
-
-            while l < r:
-                if cur + nums[l] + nums[r] == 0:
-                    res.append([cur, nums[l], nums[r]])
-                    # Drop duplicates for the second time in interation of l & r. Only used when target situation occurs, because that is the reason for dropping duplicates.
-                    while l < r - 1 and nums[l] == nums[l + 1]:
-                        l += 1
-                    while r > l + 1 and nums[r] == nums[r - 1]:
-                        r -= 1
-                if cur + nums[l] + nums[r] > 0:
-                    r -= 1
+        result = []
+        nums.sort()
+        # 找出a + b + c = 0
+        # a = nums[i], b = nums[j], c = -(a + b)
+        for i in range(len(nums)):
+            # 排序之后如果第一个元素已经大于零，那么不可能凑成三元组
+            if nums[i] > 0:
+                break
+            if i > 0 and nums[i] == nums[i - 1]: #三元组元素a去重
+                continue
+            d = {}
+            for j in range(i + 1, len(nums)):
+                if j > i + 2 and nums[j] == nums[j-1] == nums[j-2]: # 三元组元素b去重
+                    continue
+                c = 0 - (nums[i] + nums[j])
+                if c in d:
+                    result.append([nums[i], nums[j], c])
+                    d.pop(c) # 三元组元素c去重
                 else:
-                    l += 1
-        return res
+                    d[nums[j]] = j
+        return result
 ```
 
 Go：
diff --git a/problems/0018.四数之和.md b/problems/0018.四数之和.md
index 5f4c2ec9..a4d41d9b 100644
--- a/problems/0018.四数之和.md
+++ b/problems/0018.四数之和.md
@@ -207,35 +207,44 @@ class Solution {
 ```
 
 Python：
+(版本一) 双指针
 ```python
-# 双指针法
 class Solution:
     def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
-        
         nums.sort()
         n = len(nums)
-        res = []
-        for i in range(n):   
-            if i > 0 and nums[i] == nums[i - 1]: continue  # 对nums[i]去重
-            for k in range(i+1, n):
-                if k > i + 1 and nums[k] == nums[k-1]: continue  # 对nums[k]去重
-                p = k + 1
-                q = n - 1
-
-                while p < q:
-                    if nums[i] + nums[k] + nums[p] + nums[q] > target: q -= 1
-                    elif nums[i] + nums[k] + nums[p] + nums[q] < target: p += 1
+        result = []
+        for i in range(n):
+            if nums[i] > target and nums[i] > 0 and target > 0:# 剪枝（可省）
+                break
+            if i > 0 and nums[i] == nums[i-1]:# 去重
+                continue
+            for j in range(i+1, n):
+                if nums[i] + nums[j] > target and target > 0: #剪枝（可省）
+                    break
+                if j > i+1 and nums[j] == nums[j-1]: # 去重
+                    continue
+                left, right = j+1, n-1
+                while left < right:
+                    s = nums[i] + nums[j] + nums[left] + nums[right]
+                    if s == target:
+                        result.append([nums[i], nums[j], nums[left], nums[right]])
+                        while left < right and nums[left] == nums[left+1]:
+                            left += 1
+                        while left < right and nums[right] == nums[right-1]:
+                            right -= 1
+                        left += 1
+                        right -= 1
+                    elif s < target:
+                        left += 1
                     else:
-                        res.append([nums[i], nums[k], nums[p], nums[q]])
-			# 对nums[p]和nums[q]去重
-                        while p < q and nums[p] == nums[p + 1]: p += 1
-                        while p < q and nums[q] == nums[q - 1]: q -= 1
-                        p += 1
-                        q -= 1
-        return res
+                        right -= 1
+        return result
+
 ```
+(版本二) 使用字典
+
 ```python
-# 哈希表法
 class Solution(object):
     def fourSum(self, nums, target):
         """
@@ -243,36 +252,25 @@ class Solution(object):
         :type target: int
         :rtype: List[List[int]]
         """
-        # use a dict to store value:showtimes
-        hashmap = dict()
-        for n in nums:
-            if n in hashmap:
-                hashmap[n] += 1
-            else: 
-                hashmap[n] = 1
+        # 创建一个字典来存储输入列表中每个数字的频率
+        freq = {}
+        for num in nums:
+            freq[num] = freq.get(num, 0) + 1
         
-        # good thing about using python is you can use set to drop duplicates.
+        # 创建一个集合来存储最终答案，并遍历4个数字的所有唯一组合
         ans = set()
-        # ans = []  # save results by list()
         for i in range(len(nums)):
             for j in range(i + 1, len(nums)):
                 for k in range(j + 1, len(nums)):
                     val = target - (nums[i] + nums[j] + nums[k])
-                    if val in hashmap:
-                        # make sure no duplicates.
+                    if val in freq:
+                        # 确保没有重复
                         count = (nums[i] == val) + (nums[j] == val) + (nums[k] == val)
-                        if hashmap[val] > count:
-                          ans_tmp = tuple(sorted([nums[i], nums[j], nums[k], val]))
-                          ans.add(ans_tmp)
-                          # Avoiding duplication in list manner but it cause time complexity increases
-                          # if ans_tmp not in ans:  
-                          #     ans.append(ans_tmp)
-                        else:
-                            continue
-        return list(ans) 
-        # if used list() to save results, just 
-        # return ans
+                        if freq[val] > count:
+                            ans.add(tuple(sorted([nums[i], nums[j], nums[k], val])))
         
+        return [list(x) for x in ans]
+
 ```
 
 Go：
diff --git a/problems/0019.删除链表的倒数第N个节点.md b/problems/0019.删除链表的倒数第N个节点.md
index a11ff8ba..c6f5bfc7 100644
--- a/problems/0019.删除链表的倒数第N个节点.md
+++ b/problems/0019.删除链表的倒数第N个节点.md
@@ -127,21 +127,29 @@ Python:
 #     def __init__(self, val=0, next=None):
 #         self.val = val
 #         self.next = next
+
 class Solution:
     def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
-        head_dummy = ListNode()
-        head_dummy.next = head
-
-        slow, fast = head_dummy, head_dummy
-        while(n>=0): #fast先往前走n+1步
+        # 创建一个虚拟节点，并将其下一个指针设置为链表的头部
+        dummy_head = ListNode(0, head)
+        
+        # 创建两个指针，慢指针和快指针，并将它们初始化为虚拟节点
+        slow = fast = dummy_head
+        
+        # 快指针比慢指针快 n+1 步
+        for i in range(n+1):
             fast = fast.next
-            n -= 1
-        while(fast!=None):
+        
+        # 移动两个指针，直到快速指针到达链表的末尾
+        while fast:
             slow = slow.next
             fast = fast.next
-        #fast 走到结尾后，slow的下一个节点为倒数第N个节点
-        slow.next = slow.next.next #删除
-        return head_dummy.next
+        
+        # 通过更新第 (n-1) 个节点的 next 指针删除第 n 个节点
+        slow.next = slow.next.next
+        
+        return dummy_head.next
+
 ```
 Go:
 ```Go
diff --git a/problems/0024.两两交换链表中的节点.md b/problems/0024.两两交换链表中的节点.md
index 4dc051aa..2c171dde 100644
--- a/problems/0024.两两交换链表中的节点.md
+++ b/problems/0024.两两交换链表中的节点.md
@@ -186,21 +186,20 @@ Python：
 
 class Solution:
     def swapPairs(self, head: ListNode) -> ListNode:
-        res = ListNode(next=head)
-        pre = res
+        dummy_head = ListNode(next=head)
+        current = dummy_head
         
-        # 必须有pre的下一个和下下个才能交换，否则说明已经交换结束了
-        while pre.next and pre.next.next:
-            cur = pre.next
-            post = pre.next.next
+        # 必须有cur的下一个和下下个才能交换，否则说明已经交换结束了
+        while current.next and current.next.next:
+            temp = current.next # 防止节点修改
+            temp1 = current.next.next.next
             
-            # pre，cur，post对应最左，中间的，最右边的节点
-            cur.next = post.next
-            post.next = cur
-            pre.next = post
+            current.next = current.next.next
+            current.next.next = temp
+            temp.next = temp1
+            current = current.next.next
+        return dummy_head.next
 
-            pre = pre.next.next
-        return res.next
 ```
 
 Go：
diff --git a/problems/0027.移除元素.md b/problems/0027.移除元素.md
index 91150c74..90801153 100644
--- a/problems/0027.移除元素.md
+++ b/problems/0027.移除元素.md
@@ -198,6 +198,7 @@ Python：
 
 
 ``` python 3
+（版本一）快慢指针法
 class Solution:
     def removeElement(self, nums: List[int], val: int) -> int:
         # 快慢指针
@@ -213,7 +214,21 @@ class Solution:
         return slow
 ```
 
-
+``` python 3
+（版本二）暴力法
+class Solution:
+    def removeElement(self, nums: List[int], val: int) -> int:
+        i, l = 0, len(nums)
+        while i < l:
+            if nums[i] == val: # 找到等于目标值的节点
+                for j in range(i+1, l): # 移除该元素，并将后面元素向前平移
+                    nums[j - 1] = nums[j]
+                l -= 1
+                i -= 1
+            i += 1
+        return l
+                
+```
 
 
 Go：
diff --git a/problems/0028.实现strStr.md b/problems/0028.实现strStr.md
index 263c1689..86308b47 100644
--- a/problems/0028.实现strStr.md
+++ b/problems/0028.实现strStr.md
@@ -692,8 +692,67 @@ class Solution {
 ```
 
 Python3：
+（版本一）前缀表（减一）
+```python
+class Solution:
+    def getNext(self, next, s):
+        j = -1
+        next[0] = j
+        for i in range(1, len(s)):
+            while j >= 0 and s[i] != s[j+1]:
+                j = next[j]
+            if s[i] == s[j+1]:
+                j += 1
+            next[i] = j
+    
+    def strStr(self, haystack: str, needle: str) -> int:
+        if not needle:
+            return 0
+        next = [0] * len(needle)
+        self.getNext(next, needle)
+        j = -1
+        for i in range(len(haystack)):
+            while j >= 0 and haystack[i] != needle[j+1]:
+                j = next[j]
+            if haystack[i] == needle[j+1]:
+                j += 1
+            if j == len(needle) - 1:
+                return i - len(needle) + 1
+        return -1
+```
+（版本二）前缀表（不减一）
+
+```python
+class Solution:
+    def getNext(self, next: List[int], s: str) -> None:
+        j = 0
+        next[0] = 0
+        for i in range(1, len(s)):
+            while j > 0 and s[i] != s[j]:
+                j = next[j - 1]
+            if s[i] == s[j]:
+                j += 1
+            next[i] = j
+    
+    def strStr(self, haystack: str, needle: str) -> int:
+        if len(needle) == 0:
+            return 0
+        next = [0] * len(needle)
+        self.getNext(next, needle)
+        j = 0
+        for i in range(len(haystack)):
+            while j > 0 and haystack[i] != needle[j]:
+                j = next[j - 1]
+            if haystack[i] == needle[j]:
+                j += 1
+            if j == len(needle):
+                return i - len(needle) + 1
+        return -1
+```
+
+
+（版本三）暴力法
 ```python
-//暴力解法：
 class Solution(object):
     def strStr(self, haystack, needle):
         """
@@ -707,102 +766,22 @@ class Solution(object):
                 return i
         return -1    
 ```
+（版本四）使用 index
 ```python
-// 方法一
 class Solution:
     def strStr(self, haystack: str, needle: str) -> int:
-        a = len(needle)
-        b = len(haystack)
-        if a == 0:
-            return 0
-        next = self.getnext(a,needle)
-        p=-1
-        for j in range(b):
-            while p >= 0 and needle[p+1] != haystack[j]:
-                p = next[p]
-            if needle[p+1] == haystack[j]:
-                p += 1
-            if p == a-1:
-                return j-a+1
-        return -1
-
-    def getnext(self,a,needle):
-        next = ['' for i in range(a)]
-        k = -1
-        next[0] = k
-        for i in range(1, len(needle)):
-            while (k > -1 and needle[k+1] != needle[i]):
-                k = next[k]
-            if needle[k+1] == needle[i]:
-                k += 1
-            next[i] = k
-        return next
-```
-
-```python
-// 方法二
-class Solution:
-    def strStr(self, haystack: str, needle: str) -> int:
-        a = len(needle)
-        b = len(haystack)
-        if a == 0:
-            return 0
-        i = j = 0
-        next = self.getnext(a, needle)
-        while(i < b and j < a):
-            if j == -1 or needle[j] == haystack[i]:
-                i += 1
-                j += 1
-            else:
-                j = next[j]
-        if j == a:
-            return i-j
-        else:
+        try:
+            return haystack.index(needle)
+        except ValueError:
             return -1
-
-    def getnext(self, a, needle):
-        next = ['' for i in range(a)]
-        j, k = 0, -1
-        next[0] = k
-        while(j < a-1):
-            if k == -1 or needle[k] == needle[j]:
-                k += 1
-                j += 1
-                next[j] = k
-            else:
-                k = next[k]
-        return next
 ```
-
+（版本五）使用 find
 ```python
-// 前缀表（不减一）Python实现
 class Solution:
     def strStr(self, haystack: str, needle: str) -> int:
-        if len(needle) == 0:
-            return 0
-        next = self.getNext(needle)
-        j = 0
-        for i in range(len(haystack)):
-            while j >= 1 and haystack[i] != needle[j]:
-                j = next[j-1]
-            if haystack[i] == needle[j]:
-                j += 1
-            if j == len(needle):
-                return i - len(needle) + 1
-        return -1
-    
-    def getNext(self, needle):
-        next = [0] * len(needle)
-        j = 0
-        next[0] = j
-        for i in range(1, len(needle)):
-            while j >= 1 and needle[i] != needle[j]:
-                j = next[j-1]
-            if needle[i] == needle[j]:
-                j += 1
-            next[i] = j
-        return next
-```
+        return haystack.find(needle)
+	
+```       
 
 Go：
 
diff --git a/problems/0084.柱状图中最大的矩形.md b/problems/0084.柱状图中最大的矩形.md
index eb064143..f9a83508 100644
--- a/problems/0084.柱状图中最大的矩形.md
+++ b/problems/0084.柱状图中最大的矩形.md
@@ -307,6 +307,33 @@ class Solution {
     }
 }
 ```
+单调栈精简
+```java
+class Solution {
+    public int largestRectangleArea(int[] heights) {
+        int[] newHeight = new int[heights.length + 2];
+        System.arraycopy(heights, 0, newHeight, 1, heights.length);
+        newHeight[heights.length+1] = 0;
+        newHeight[0] = 0;
+
+        Stack<Integer> stack = new Stack<>();
+        stack.push(0);
+
+        int res = 0;
+        for (int i = 1; i < newHeight.length; i++) {
+            while (newHeight[i] < newHeight[stack.peek()]) {
+                int mid = stack.pop();
+                int w = i - stack.peek() - 1;
+                int h = newHeight[mid];
+                res = Math.max(res, w * h);
+            }
+            stack.push(i);
+
+        }
+        return res;
+    }
+}
+```
 
 Python3:
 
diff --git a/problems/0101.对称二叉树.md b/problems/0101.对称二叉树.md
index b75e9ff2..81ca79a2 100644
--- a/problems/0101.对称二叉树.md
+++ b/problems/0101.对称二叉树.md
@@ -442,25 +442,31 @@ class Solution:
 层次遍历
 ```python
 class Solution:
-    def isSymmetric(self, root: Optional[TreeNode]) -> bool:
+    def isSymmetric(self, root: TreeNode) -> bool:
         if not root:
             return True
-
-        que = [root]
-        while que:
-            this_level_length = len(que)
-            for i in range(this_level_length // 2):
-                # 要么其中一个是None但另外一个不是
-                if (not que[i] and que[this_level_length - 1 - i]) or (que[i] and not que[this_level_length - 1 - i]):
-                    return False
-                # 要么两个都不是None
-                if que[i] and que[i].val != que[this_level_length - 1 - i].val:
-                    return False
-            for i in range(this_level_length):
-                if not que[i]: continue
-                que.append(que[i].left)
-                que.append(que[i].right)
-            que = que[this_level_length:]
+        
+        queue = collections.deque([root.left, root.right])
+        
+        while queue:
+            level_size = len(queue)
+            
+            if level_size % 2 != 0:
+                return False
+            
+            level_vals = []
+            for i in range(level_size):
+                node = queue.popleft()
+                if node:
+                    level_vals.append(node.val)
+                    queue.append(node.left)
+                    queue.append(node.right)
+                else:
+                    level_vals.append(None)
+                    
+            if level_vals != level_vals[::-1]:
+                return False
+            
         return True
 ```
 
diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md
index 26f0ae2d..c2ad9508 100644
--- a/problems/0102.二叉树的层序遍历.md
+++ b/problems/0102.二叉树的层序遍历.md
@@ -171,47 +171,59 @@ python3代码：
 
 
 ```python
+# 利用长度法
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
 class Solution:
-    """二叉树层序遍历迭代解法"""
-
-    def levelOrder(self, root: TreeNode) -> List[List[int]]:
-        results = []
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
         if not root:
-            return results
-
-        from collections import deque
-        que = deque([root])
-
-        while que:
-            size = len(que)
-            result = []
-            for _ in range(size):
-                cur = que.popleft()
-                result.append(cur.val)
+            return []
+        queue = collections.deque([root])
+        result = []
+        while queue:
+            level = []
+            for _ in range(len(queue)):
+                cur = queue.popleft()
+                level.append(cur.val)
                 if cur.left:
-                    que.append(cur.left)
+                    queue.append(cur.left)
                 if cur.right:
-                    que.append(cur.right)
-            results.append(result)
-
-        return results
+                    queue.append(cur.right)
+            result.append(level)
+        return result
 ```
-
 ```python
 # 递归法
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
 class Solution:
-    def levelOrder(self, root: TreeNode) -> List[List[int]]:
-        res = []
-        def helper(root, depth):
-            if not root: return []
-            if len(res) == depth: res.append([]) # start the current depth
-            res[depth].append(root.val) # fulfil the current depth
-            if  root.left: helper(root.left, depth + 1) # process child nodes for the next depth
-            if  root.right: helper(root.right, depth + 1)
-        helper(root, 0)
-        return res
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
+        levels = []
+        self.helper(root, 0, levels)
+        return levels
+    
+    def helper(self, node, level, levels):
+        if not node:
+            return
+        if len(levels) == level:
+            levels.append([])
+        levels[level].append(node.val)
+        self.helper(node.left, level + 1, levels)
+        self.helper(node.right, level + 1, levels)
+
+
 ```
 
+
+
 go:
 
 ```go
@@ -500,27 +512,29 @@ python代码：
 class Solution:
     """二叉树层序遍历II迭代解法"""
 
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
     def levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
-        results = []
         if not root:
-            return results
-
-        from collections import deque
-        que = deque([root])
-
-        while que:
-            result = []
-            for _ in range(len(que)):
-                cur = que.popleft()
-                result.append(cur.val)
+            return []
+        queue = collections.deque([root])
+        result = []
+        while queue:
+            level = []
+            for _ in range(len(queue)):
+                cur = queue.popleft()
+                level.append(cur.val)
                 if cur.left:
-                    que.append(cur.left)
+                    queue.append(cur.left)
                 if cur.right:
-                    que.append(cur.right)
-            results.append(result)
-
-        results.reverse()
-        return results
+                    queue.append(cur.right)
+            result.append(level)
+        return result[::-1]
 ```
 
 Java：
@@ -821,35 +835,35 @@ public:
 python代码：
 
 ```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
 class Solution:
     def rightSideView(self, root: TreeNode) -> List[int]:
         if not root:
             return []
-
-        # deque来自collections模块，不在力扣平台时，需要手动写入
-        # 'from collections import deque' 导入
-        # deque相比list的好处是，list的pop(0)是O(n)复杂度，deque的popleft()是O(1)复杂度
-
-        quene = deque([root])
-        out_list = []
-
-        while quene:
-            # 每次都取最后一个node就可以了
-            node = quene[-1]
-            out_list.append(node.val)
-
-            # 执行这个遍历的目的是获取下一层所有的node
-            for _ in range(len(quene)):
-                node = quene.popleft()
+        
+        queue = collections.deque([root])
+        right_view = []
+        
+        while queue:
+            level_size = len(queue)
+            
+            for i in range(level_size):
+                node = queue.popleft()
+                
+                if i == level_size - 1:
+                    right_view.append(node.val)
+                
                 if node.left:
-                    quene.append(node.left)
+                    queue.append(node.left)
                 if node.right:
-                    quene.append(node.right)
-
-        return out_list
-
-# 执行用时：36 ms, 在所有 Python3 提交中击败了89.47%的用户
-# 内存消耗：14.6 MB, 在所有 Python3 提交中击败了96.65%的用户
+                    queue.append(node.right)
+        
+        return right_view
 ```
 
 
@@ -1107,27 +1121,38 @@ python代码：
 class Solution:
     """二叉树层平均值迭代解法"""
 
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
     def averageOfLevels(self, root: TreeNode) -> List[float]:
-        results = []
         if not root:
-            return results
+            return []
 
-        from collections import deque
-        que = deque([root])
-
-        while que:
-            size = len(que)
-            sum_ = 0
-            for _ in range(size):
-                cur = que.popleft()
-                sum_ += cur.val
-                if cur.left:
-                    que.append(cur.left)
-                if cur.right:
-                    que.append(cur.right)
-            results.append(sum_ / size)
-
-        return results
+        queue = collections.deque([root])
+        averages = []
+        
+        while queue:
+            size = len(queue)
+            level_sum = 0
+            
+            for i in range(size):
+                node = queue.popleft()
+                
+                
+                level_sum += node.val
+                    
+                if node.left:
+                    queue.append(node.left)
+                if node.right:
+                    queue.append(node.right)
+            
+            averages.append(level_sum / size)
+        
+        return averages
 ```
 
 java:
@@ -1401,28 +1426,36 @@ public:
 python代码：
 
 ```python
+"""
+# Definition for a Node.
+class Node:
+    def __init__(self, val=None, children=None):
+        self.val = val
+        self.children = children
+"""
+
 class Solution:
-    """N叉树的层序遍历迭代法"""
-
     def levelOrder(self, root: 'Node') -> List[List[int]]:
-        results = []
         if not root:
-            return results
+            return []
 
-        from collections import deque
-        que = deque([root])
+        result = []
+        queue = collections.deque([root])
 
-        while que:
-            result = []
-            for _ in range(len(que)):
-                cur = que.popleft()
-                result.append(cur.val)
-                # cur.children 是 Node 对象组成的列表，也可能为 None
-                if cur.children:
-                    que.extend(cur.children)
-            results.append(result)
+        while queue:
+            level_size = len(queue)
+            level = []
 
-        return results
+            for _ in range(level_size):
+                node = queue.popleft()
+                level.append(node.val)
+
+                for child in node.children:
+                    queue.append(child)
+
+            result.append(level)
+
+        return result
 ```
 
 ```python
@@ -1728,22 +1761,37 @@ public:
 python代码：
 
 ```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
 class Solution:
     def largestValues(self, root: TreeNode) -> List[int]:
-        if root is None:
+        if not root:
             return []
-        queue = [root]
-        out_list = []
+
+        result = []
+        queue = collections.deque([root])
+
         while queue:
-            length = len(queue)
-            in_list = []
-            for _ in range(length):
-                curnode = queue.pop(0)
-                in_list.append(curnode.val)
-                if curnode.left: queue.append(curnode.left)
-                if curnode.right: queue.append(curnode.right)
-            out_list.append(max(in_list))
-        return out_list
+            level_size = len(queue)
+            max_val = float('-inf')
+
+            for _ in range(level_size):
+                node = queue.popleft()
+                max_val = max(max_val, node.val)
+
+                if node.left:
+                    queue.append(node.left)
+
+                if node.right:
+                    queue.append(node.right)
+
+            result.append(max_val)
+
+        return result
 ```
 
 java代码：
@@ -2048,36 +2096,40 @@ class Solution {
 python代码：
 
 ```python
-# 层序遍历解法
+"""
+# Definition for a Node.
+class Node:
+    def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
+        self.val = val
+        self.left = left
+        self.right = right
+        self.next = next
+"""
 class Solution:
     def connect(self, root: 'Node') -> 'Node':
         if not root:
-            return None
-        queue = [root]
+            return root
+        
+        queue = collections.deque([root])
+        
         while queue:
-            n = len(queue)
-            for i in range(n):
-                node = queue.pop(0)
+            level_size = len(queue)
+            prev = None
+            
+            for i in range(level_size):
+                node = queue.popleft()
+                
+                if prev:
+                    prev.next = node
+                
+                prev = node
+                
                 if node.left:
                     queue.append(node.left)
+                
                 if node.right:
                     queue.append(node.right)
-                if i == n - 1:
-                    break
-                node.next = queue[0]
-        return root
-
-# 链表解法
-class Solution:
-    def connect(self, root: 'Node') -> 'Node':
-        first = root
-        while first:
-            cur = first
-            while cur:  # 遍历每一层的节点
-                if cur.left: cur.left.next = cur.right  # 找左节点的next
-                if cur.right and cur.next: cur.right.next = cur.next.left  # 找右节点的next
-                cur = cur.next # cur同层移动到下一节点
-            first = first.left  # 从本层扩展到下一层
+        
         return root
 ```
 
@@ -2329,21 +2381,41 @@ python代码：
 
 ```python
 # 层序遍历解法
+"""
+# Definition for a Node.
+class Node:
+    def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
+        self.val = val
+        self.left = left
+        self.right = right
+        self.next = next
+"""
+
 class Solution:
     def connect(self, root: 'Node') -> 'Node':
         if not root:
-            return None
-        queue = [root]
-        while queue:  # 遍历每一层
-            length = len(queue)
-            tail = None # 每一层维护一个尾节点
-            for i in range(length): # 遍历当前层
-                curnode = queue.pop(0)
-                if tail:
-                    tail.next = curnode # 让尾节点指向当前节点
-                tail = curnode # 让当前节点成为尾节点
-                if curnode.left : queue.append(curnode.left)
-                if curnode.right: queue.append(curnode.right)
+            return root
+        
+        queue = collections.deque([root])
+        
+        while queue:
+            level_size = len(queue)
+            prev = None
+            
+            for i in range(level_size):
+                node = queue.popleft()
+                
+                if prev:
+                    prev.next = node
+                
+                prev = node
+                
+                if node.left:
+                    queue.append(node.left)
+                
+                if node.right:
+                    queue.append(node.right)
+        
         return root
 
 ```
@@ -2592,24 +2664,31 @@ class Solution {
 Python：
 
 ```python 3
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
 class Solution:
     def maxDepth(self, root: TreeNode) -> int:
-        if root == None:
+        if not root:
             return 0
-
-        queue_ = [root]
+        
         depth = 0
-        while queue_:
-            length = len(queue_)
-            for i in range(length):
-                cur = queue_.pop(0)
-                sub.append(cur.val)
-                #子节点入队列
-                if cur.left: queue_.append(cur.left)
-                if cur.right: queue_.append(cur.right)
+        queue = collections.deque([root])
+        
+        while queue:
             depth += 1
-
+            for _ in range(len(queue)):
+                node = queue.popleft()
+                if node.left:
+                    queue.append(node.left)
+                if node.right:
+                    queue.append(node.right)
+        
         return depth
+
 ```
 
 Go：
@@ -2859,23 +2938,26 @@ Python 3：
 #         self.right = right
 class Solution:
     def minDepth(self, root: TreeNode) -> int:
-        if root == None:
+        if not root:
             return 0
+        depth = 0
+        queue = collections.deque([root])
+        
+        while queue:
+            depth += 1 
+            for _ in range(len(queue)):
+                node = queue.popleft()
+                
+                if not node.left and not node.right:
+                    return depth
+            
+                if node.left:
+                    queue.append(node.left)
+                    
+                if node.right:
+                    queue.append(node.right)
 
-        #根节点的深度为1
-        queue_ = [(root,1)]
-        while queue_:
-            cur, depth = queue_.pop(0)
-
-            if cur.left == None and cur.right == None:
-                return depth
-            #先左子节点，由于左子节点没有孩子，则就是这一层了
-            if cur.left:
-                queue_.append((cur.left,depth + 1))
-            if cur.right:
-                queue_.append((cur.right,depth + 1))
-
-        return 0
+        return depth
 ```
 
 Go：
diff --git a/problems/0104.二叉树的最大深度.md b/problems/0104.二叉树的最大深度.md
index 96169b32..7130867b 100644
--- a/problems/0104.二叉树的最大深度.md
+++ b/problems/0104.二叉树的最大深度.md
@@ -419,86 +419,107 @@ class solution:
         return 1 + max(self.maxdepth(root.left), self.maxdepth(root.right))
 ```
 
-迭代法：
+层序遍历迭代法：
 ```python
-import collections
-class solution:
-    def maxdepth(self, root: treenode) -> int:
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def maxDepth(self, root: TreeNode) -> int:
         if not root:
             return 0
-        depth = 0 #记录深度
-        queue = collections.deque()
-        queue.append(root)
+        
+        depth = 0
+        queue = collections.deque([root])
+        
         while queue:
-            size = len(queue)
             depth += 1
-            for i in range(size):
+            for _ in range(len(queue)):
                 node = queue.popleft()
                 if node.left:
                     queue.append(node.left)
                 if node.right:
                     queue.append(node.right)
+        
         return depth
+
 ```
 
 ### 559.n叉树的最大深度 
 
 递归法：
 ```python
-class solution:
-    def maxdepth(self, root: 'node') -> int:
+class Solution:
+    def maxDepth(self, root: 'Node') -> int:
         if not root:
             return 0
-        depth = 0
-        for i in range(len(root.children)):
-            depth = max(depth, self.maxdepth(root.children[i]))
-        return depth + 1
+        
+        max_depth = 1
+        
+        for child in root.children:
+            max_depth = max(max_depth, self.maxDepth(child) + 1)
+        
+        return max_depth
 ```
 
 迭代法：
 ```python
-import collections
-class solution:
-    def maxdepth(self, root: 'node') -> int:
-        queue = collections.deque()
-        if root:
-            queue.append(root)
-        depth = 0 #记录深度
+"""
+# Definition for a Node.
+class Node:
+    def __init__(self, val=None, children=None):
+        self.val = val
+        self.children = children
+"""
+
+class Solution:
+    def maxDepth(self, root: TreeNode) -> int:
+        if not root:
+            return 0
+        
+        depth = 0
+        queue = collections.deque([root])
+        
         while queue:
-            size = len(queue)
             depth += 1
-            for i in range(size):
+            for _ in range(len(queue)):
                 node = queue.popleft()
-                for j in range(len(node.children)):
-                    if node.children[j]:
-                        queue.append(node.children[j])
+                for child in node.children:
+                    queue.append(child)
+        
         return depth
+
 ```
 
-使用栈来模拟后序遍历依然可以
+使用栈
 ```python
-class solution:
-    def maxdepth(self, root: 'node') -> int:
-        st = []
-        if root:
-            st.append(root)
-        depth = 0
-        result = 0
-        while st:
-            node = st.pop()
-            if node != none:
-                st.append(node) #中
-                st.append(none)
-                depth += 1
-                for i in range(len(node.children)): #处理孩子
-                    if node.children[i]:
-                        st.append(node.children[i])
-                    
-            else:
-                node = st.pop()
-                depth -= 1
-            result = max(result, depth)
-        return result
+"""
+# Definition for a Node.
+class Node:
+    def __init__(self, val=None, children=None):
+        self.val = val
+        self.children = children
+"""
+
+class Solution:
+    def maxDepth(self, root: 'Node') -> int:
+        if not root:
+            return 0
+        
+        max_depth = 0
+        
+        stack = [(root, 1)]
+        
+        while stack:
+            node, depth = stack.pop()
+            max_depth = max(max_depth, depth)
+            for child in node.children:
+                stack.append((child, depth + 1))
+        
+        return max_depth
 ```
 
 
diff --git a/problems/0106.从中序与后序遍历序列构造二叉树.md b/problems/0106.从中序与后序遍历序列构造二叉树.md
index adb374f9..8fc973e0 100644
--- a/problems/0106.从中序与后序遍历序列构造二叉树.md
+++ b/problems/0106.从中序与后序遍历序列构造二叉树.md
@@ -622,7 +622,42 @@ class Solution {
     }
 }
 ```
+```java
+class Solution {
+    public TreeNode buildTree(int[] inorder, int[] postorder) {
+        if(postorder.length == 0 || inorder.length == 0)
+            return null;
+        return buildHelper(inorder, 0, inorder.length, postorder, 0, postorder.length);
+    
+    }
+    private TreeNode buildHelper(int[] inorder, int inorderStart, int inorderEnd, int[] postorder, int postorderStart, int postorderEnd){
+        if(postorderStart == postorderEnd)
+            return null;
+        int rootVal = postorder[postorderEnd - 1];
+        TreeNode root = new TreeNode(rootVal);
+        int middleIndex;
+        for (middleIndex = inorderStart; middleIndex < inorderEnd; middleIndex++){
+            if(inorder[middleIndex] == rootVal)
+                break;
+        }
 
+        int leftInorderStart = inorderStart; 
+        int leftInorderEnd = middleIndex;
+        int rightInorderStart = middleIndex + 1;
+        int rightInorderEnd = inorderEnd;
+
+
+        int leftPostorderStart = postorderStart;
+        int leftPostorderEnd = postorderStart + (middleIndex - inorderStart);
+        int rightPostorderStart = leftPostorderEnd;
+        int rightPostorderEnd = postorderEnd - 1;
+        root.left = buildHelper(inorder, leftInorderStart, leftInorderEnd,  postorder, leftPostorderStart, leftPostorderEnd);
+        root.right = buildHelper(inorder, rightInorderStart, rightInorderEnd, postorder, rightPostorderStart, rightPostorderEnd);
+
+        return root;
+    }  
+}
+```
 105.从前序与中序遍历序列构造二叉树
 
 ```java
diff --git a/problems/0110.平衡二叉树.md b/problems/0110.平衡二叉树.md
index 804c95eb..a3bc77fb 100644
--- a/problems/0110.平衡二叉树.md
+++ b/problems/0110.平衡二叉树.md
@@ -532,6 +532,24 @@ class Solution:
         else:
             return 1 + max(left_height, right_height)
 ```
+递归法精简版：
+
+```python
+class Solution:
+    def isBalanced(self, root: TreeNode) -> bool:
+        return self.height(root) != -1
+    def height(self, node: TreeNode) -> int:
+            if not node:
+                return 0
+            left = self.height(node.left)
+            if left == -1:
+                return -1
+            right = self.height(node.right)
+            if right == -1 or abs(left - right) > 1:
+                return -1
+            return max(left, right) + 1
+```
+
 
 迭代法：
 
diff --git a/problems/0111.二叉树的最小深度.md b/problems/0111.二叉树的最小深度.md
index 47569b05..d0bdf4be 100644
--- a/problems/0111.二叉树的最小深度.md
+++ b/problems/0111.二叉树的最小深度.md
@@ -305,46 +305,98 @@ class Solution {
 递归法：
 
 ```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
 class Solution:
     def minDepth(self, root: TreeNode) -> int:
         if not root:
             return 0
+        
         if not root.left and not root.right:
             return 1
-
-        min_depth = 10**9
+        
+        left_depth = float('inf')
+        right_depth = float('inf')
+        
         if root.left:
-            min_depth = min(self.minDepth(root.left), min_depth) # 获得左子树的最小高度
+            left_depth = self.minDepth(root.left)
         if root.right:
-            min_depth = min(self.minDepth(root.right), min_depth) # 获得右子树的最小高度
-        return min_depth + 1
+            right_depth = self.minDepth(root.right)
+        
+        return 1 + min(left_depth, right_depth)
+
 ```
 
 迭代法：
 
 ```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
 class Solution:
     def minDepth(self, root: TreeNode) -> int:
         if not root:
             return 0
-        que = deque()
-        que.append(root)
-        res = 1
-
-        while que:
-            for _ in range(len(que)):
-                node = que.popleft()
-                # 当左右孩子都为空的时候，说明是最低点的一层了，退出
+        depth = 0
+        queue = collections.deque([root])
+        
+        while queue:
+            depth += 1 
+            for _ in range(len(queue)):
+                node = queue.popleft()
+                
                 if not node.left and not node.right:
-                    return res
-                if node.left is not None:
-                    que.append(node.left)
-                if node.right is not None:
-                    que.append(node.right)
-            res += 1
-        return res
+                    return depth
+            
+                if node.left:
+                    queue.append(node.left)
+                    
+                if node.right:
+                    queue.append(node.right)
+
+        return depth
 ```
 
+迭代法：
+
+```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+
+class Solution:
+    def minDepth(self, root: TreeNode) -> int:
+        if not root:
+            return 0
+        
+        queue = collections.deque([(root, 1)])
+        
+        while queue:
+            node, depth = queue.popleft()
+            
+            # Check if the node is a leaf node
+            if not node.left and not node.right:
+                return depth
+            
+            # Add left and right child to the queue
+            if node.left:
+                queue.append((node.left, depth+1))
+            if node.right:
+                queue.append((node.right, depth+1))
+        
+        return 0
+
+```
 
 ## Go
 
diff --git a/problems/0112.路径总和.md b/problems/0112.路径总和.md
index 51aa1956..0a4f5714 100644
--- a/problems/0112.路径总和.md
+++ b/problems/0112.路径总和.md
@@ -487,140 +487,191 @@ class Solution {
 
 ### 0112.路径总和
 
-**递归**
-
+(版本一) 递归
 ```python
-class solution:
-    def haspathsum(self, root: treenode, targetsum: int) -> bool:
-        def isornot(root, targetsum) -> bool:
-            if (not root.left) and (not root.right) and targetsum == 0:
-                return true  # 遇到叶子节点，并且计数为0
-            if (not root.left) and (not root.right):
-                return false  # 遇到叶子节点，计数不为0
-            if root.left:
-                targetsum -= root.left.val  # 左节点
-                if isornot(root.left, targetsum): return true  # 递归，处理左节点
-                targetsum += root.left.val  # 回溯
-            if root.right:
-                targetsum -= root.right.val  # 右节点
-                if isornot(root.right, targetsum): return true  # 递归，处理右节点
-                targetsum += root.right.val  # 回溯
-            return false
-
-        if root == none:
-            return false  # 别忘记处理空treenode
-        else:
-            return isornot(root, targetsum - root.val)
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def traversal(self, cur: TreeNode, count: int) -> bool:
+        if not cur.left and not cur.right and count == 0: # 遇到叶子节点，并且计数为0
+            return True
+        if not cur.left and not cur.right: # 遇到叶子节点直接返回
+            return False
+        
+        if cur.left: # 左
+            count -= cur.left.val
+            if self.traversal(cur.left, count): # 递归，处理节点
+                return True
+            count += cur.left.val # 回溯，撤销处理结果
             
-class Solution: # 简洁版
-    def hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:
-        if not root: return False
-        if root.left==root.right==None and root.val == targetSum: return True
-        return self.hasPathSum(root.left,targetSum-root.val) or self.hasPathSum(root.right,targetSum-root.val)          
+        if cur.right: # 右
+            count -= cur.right.val
+            if self.traversal(cur.right, count): # 递归，处理节点
+                return True
+            count += cur.right.val # 回溯，撤销处理结果
+            
+        return False
+    
+    def hasPathSum(self, root: TreeNode, sum: int) -> bool:
+        if root is None:
+            return False
+        return self.traversal(root, sum - root.val)      
 ```
 
-**迭代 - 层序遍历**
-
+(版本二) 递归 + 精简
 ```python
-class solution:
-    def haspathsum(self, root: treenode, targetsum: int) -> bool:
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def hasPathSum(self, root: TreeNode, sum: int) -> bool:
         if not root:
-            return false
-
-        stack = []  # [(当前节点，路径数值), ...]
-        stack.append((root, root.val))
-
-        while stack:
-            cur_node, path_sum = stack.pop()
-
-            if not cur_node.left and not cur_node.right and path_sum == targetsum:
-                return true
-
-            if cur_node.right:
-                stack.append((cur_node.right, path_sum + cur_node.right.val))
-
-            if cur_node.left:
-                stack.append((cur_node.left, path_sum + cur_node.left.val))
-
-        return false
+            return False
+        if not root.left and not root.right and sum == root.val:
+            return True
+        return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val)
+  
 ```
+(版本三) 迭代
+```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def hasPathSum(self, root: TreeNode, sum: int) -> bool:
+        if not root:
+            return False
+        # 此时栈里要放的是pair<节点指针，路径数值>
+        st = [(root, root.val)]
+        while st:
+            node, path_sum = st.pop()
+            # 如果该节点是叶子节点了，同时该节点的路径数值等于sum，那么就返回true
+            if not node.left and not node.right and path_sum == sum:
+                return True
+            # 右节点，压进去一个节点的时候，将该节点的路径数值也记录下来
+            if node.right:
+                st.append((node.right, path_sum + node.right.val))
+            # 左节点，压进去一个节点的时候，将该节点的路径数值也记录下来
+            if node.left:
+                st.append((node.left, path_sum + node.left.val))
+        return False
+
+
+  
+```
+
+
 
 ### 0113.路径总和-ii
 
-**递归**
-
+(版本一) 递归
 ```python
-class solution:
-    def pathsum(self, root: treenode, targetsum: int) -> list[list[int]]:
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def __init__(self):
+        self.result = []
+        self.path = []
 
-        def traversal(cur_node, remain):
-            if not cur_node.left and not cur_node.right:
-                if remain == 0:
-                    result.append(path[:])
-                return
+    def traversal(self, cur, count):
+        if not cur.left and not cur.right and count == 0: # 遇到了叶子节点且找到了和为sum的路径
+            self.result.append(self.path[:])
+            return
 
-            if cur_node.left:
-                path.append(cur_node.left.val)
-                traversal(cur_node.left, remain-cur_node.left.val)
-                path.pop()
+        if not cur.left and not cur.right: # 遇到叶子节点而没有找到合适的边，直接返回
+            return
 
-            if cur_node.right:
-                path.append(cur_node.right.val)
-                traversal(cur_node.right, remain-cur_node.right.val)
-                path.pop()
+        if cur.left: # 左 （空节点不遍历）
+            self.path.append(cur.left.val)
+            count -= cur.left.val
+            self.traversal(cur.left, count) # 递归
+            count += cur.left.val # 回溯
+            self.path.pop() # 回溯
 
-        result, path = [], []
+        if cur.right: #  右 （空节点不遍历）
+            self.path.append(cur.right.val) 
+            count -= cur.right.val
+            self.traversal(cur.right, count) # 递归
+            count += cur.right.val # 回溯
+            self.path.pop() # 回溯
+
+        return
+
+    def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
+        self.result.clear()
+        self.path.clear()
         if not root:
-            return []
-        path.append(root.val)
-        traversal(root, targetsum - root.val)
-        return result
+            return self.result
+        self.path.append(root.val) # 把根节点放进路径
+        self.traversal(root, sum - root.val)
+        return self.result 
 ```
 
-**迭代法，用第二个队列保存目前的总和与路径**
-
+(版本二) 递归 + 精简
 ```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
 class Solution:
-    def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
-        if not root:
-            return []
-        que, temp = deque([root]), deque([(root.val, [root.val])])
+    def pathSum(self, root: TreeNode, targetSum: int) -> List[List[int]]:
+        
         result = []
-        while que:
-            for _ in range(len(que)):
-                node = que.popleft()
-                value, path = temp.popleft()
-                if (not node.left) and (not node.right):
-                    if value == targetSum:
-                        result.append(path)
-                if node.left:
-                    que.append(node.left)
-                    temp.append((node.left.val+value, path+[node.left.val]))
-                if node.right:
-                    que.append(node.right)
-                    temp.append((node.right.val+value, path+[node.right.val]))
+        self.traversal(root, targetSum, [], result)
         return result
+    def traversal(self,node, count, path, result):
+            if not node:
+                return
+            path.append(node.val)
+            count -= node.val
+            if not node.left and not node.right and count == 0:
+                result.append(list(path))
+            self.traversal(node.left, count, path, result)
+            self.traversal(node.right, count, path, result)
+            path.pop()
 ```
-
-**迭代法，前序遍历**
-
+(版本三) 迭代
 ```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
 class Solution:
-    def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
-        if not root: return []
-        stack, path_stack,result = [[root,root.val]],[[root.val]],[]
+    def pathSum(self, root: TreeNode, targetSum: int) -> List[List[int]]:
+        if not root:
+            return []
+        stack = [(root, [root.val])]
+        res = []
         while stack:
-            cur,cursum = stack.pop()
-            path = path_stack.pop()
-            if cur.left==cur.right==None:
-                if cursum==targetSum: result.append(path)
-            if cur.right:
-                stack.append([cur.right,cursum+cur.right.val])
-                path_stack.append(path+[cur.right.val])
-            if cur.left: 
-                stack.append([cur.left,cursum+cur.left.val])
-                path_stack.append(path+[cur.left.val])
-        return result
+            node, path = stack.pop()
+            if not node.left and not node.right and sum(path) == targetSum:
+                res.append(path)
+            if node.right:
+                stack.append((node.right, path + [node.right.val]))
+            if node.left:
+                stack.append((node.left, path + [node.left.val]))
+        return res
+
+
+  
 ```
 ## go
 
diff --git a/problems/0142.环形链表II.md b/problems/0142.环形链表II.md
index 46df4777..e80a715a 100644
--- a/problems/0142.环形链表II.md
+++ b/problems/0142.环形链表II.md
@@ -221,25 +221,55 @@ public class Solution {
 Python：
 
 ```python
+（版本一）快慢指针法
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+
 class Solution:
     def detectCycle(self, head: ListNode) -> ListNode:
-        slow, fast = head, head
+        slow = head
+        fast = head
+        
         while fast and fast.next:
             slow = slow.next
             fast = fast.next.next
-            # 如果相遇
+            
+            # If there is a cycle, the slow and fast pointers will eventually meet
             if slow == fast:
-                p = head
-                q = slow
-                while p!=q:
-                    p = p.next
-                    q = q.next
-                #你也可以return q
-                return p
-
+                # Move one of the pointers back to the start of the list
+                slow = head
+                while slow != fast:
+                    slow = slow.next
+                    fast = fast.next
+                return slow
+        # If there is no cycle, return None
         return None
 ```
+```python
+（版本二）集合法
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
 
+
+class Solution:
+    def detectCycle(self, head: ListNode) -> ListNode:
+        visited = set()
+        
+        while head:
+            if head in visited:
+                return head
+            visited.add(head)
+            head = head.next
+        
+        return None
+```
 Go：
 
 ```go
diff --git a/problems/0151.翻转字符串里的单词.md b/problems/0151.翻转字符串里的单词.md
index 8fa7c77c..4474f1c6 100644
--- a/problems/0151.翻转字符串里的单词.md
+++ b/problems/0151.翻转字符串里的单词.md
@@ -434,134 +434,40 @@ class Solution {
 ```
 
 python:
-
+（版本一）先删除空白，然后整个反转，最后单词反转。
+**因为字符串是不可变类型，所以反转单词的时候，需要将其转换成列表，然后通过join函数再将其转换成列表，所以空间复杂度不是O(1)**
 ```Python
 class Solution:
-  #1.去除多余的空格
-        def trim_spaces(self, s):     
-            n = len(s)
-            left = 0
-            right = n-1
-        
-            while left <= right and s[left] == ' ':    #去除开头的空格
-                left += 1
-            while left <= right and s[right] == ' ':   #去除结尾的空格
-                right -= 1
-            tmp = []
-            while left <= right:                      #去除单词中间多余的空格
-                if s[left] != ' ':
-                    tmp.append(s[left])
-                elif tmp[-1] != ' ':                 #当前位置是空格，但是相邻的上一个位置不是空格，则该空格是合理的
-                    tmp.append(s[left])
-                left += 1
-            return tmp
-	    
- #2.翻转字符数组
-        def reverse_string(self, nums, left, right):
-            while left < right:
-                nums[left], nums[right] = nums[right], nums[left]
-                left += 1
-                right -= 1
-            return None
-	    
- #3.翻转每个单词
-        def reverse_each_word(self, nums):
-            start = 0
-            end = 0
-            n = len(nums)
-            while start < n:
-                while end < n and nums[end] != ' ':
-                    end += 1
-                self.reverse_string(nums, start, end-1)
-                start = end + 1
-                end += 1
-            return None
-
-#4.翻转字符串里的单词
-        def reverseWords(self, s):                #测试用例："the sky is blue"
-            l = self.trim_spaces(s)               #输出：['t', 'h', 'e', ' ', 's', 'k', 'y', ' ', 'i', 's', ' ', 'b', 'l', 'u', 'e'
-            self.reverse_string(l,  0, len(l)-1)  #输出：['e', 'u', 'l', 'b', ' ', 's', 'i', ' ', 'y', 'k', 's', ' ', 'e', 'h', 't']
-            self.reverse_each_word(l)             #输出：['b', 'l', 'u', 'e', ' ', 'i', 's', ' ', 's', 'k', 'y', ' ', 't', 'h', 'e']
-            return ''.join(l)                    #输出：blue is sky the
-
+    def reverseWords(self, s: str) -> str:
+        # 删除前后空白
+        s = s.strip()
+        # 反转整个字符串
+        s = s[::-1]
+        # 将字符串拆分为单词，并反转每个单词
+        s = ' '.join(word[::-1] for word in s.split())
+        return s
 
 ```
+（版本二）使用双指针
 
 ```python
 class Solution:
     def reverseWords(self, s: str) -> str:
-        # method 1 - Rude but work & efficient method.
-        s_list = [i for i in s.split(" ") if len(i) > 0]
-        return " ".join(s_list[::-1])
+        # 将字符串拆分为单词，即转换成列表类型
+        words = s.split()
 
-        # method 2 - Carlo's idea
-        def trim_head_tail_space(ss: str):
-            p = 0
-            while p < len(ss) and ss[p] == " ":
-                p += 1
-            return ss[p:]
+        # 反转单词
+        left, right = 0, len(words) - 1
+        while left < right:
+            words[left], words[right] = words[right], words[left]
+            left += 1
+            right -= 1
 
-        # Trim the head and tail space
-        s = trim_head_tail_space(s)
-        s = trim_head_tail_space(s[::-1])[::-1]
-
-        pf, ps, s = 0, 0, s[::-1] # Reverse the string.
-        while pf < len(s):
-            if s[pf] == " ":
-                # Will not excede. Because we have clean the tail space.
-                if s[pf] == s[pf + 1]:
-                    s = s[:pf] + s[pf + 1:]
-                    continue
-                else:
-                    s = s[:ps] + s[ps: pf][::-1] + s[pf:]
-                    ps, pf = pf + 1, pf + 2
-            else:
-                pf += 1
-        return s[:ps] + s[ps:][::-1] # Must do the last step, because the last word is omit though the pointers are on the correct positions,
+        # 将列表转换成字符串
+        return " ".join(words)
 ```
 
-```python
-class Solution: # 使用双指针法移除空格
-    def reverseWords(self, s: str) -> str:
-        
-        def removeextraspace(s):
-            start = 0; end = len(s)-1
-            while s[start]==' ':
-                start+=1
-            while s[end]==' ':
-                end-=1
-            news = list(s[start:end+1])
-            slow = fast = 0
-            while fast<len(news):
-                while fast>0 and news[fast]==news[fast-1]==' ':
-                    fast+=1
-                news[slow]=news[fast]
-                slow+=1; fast+=1
-            #return "".join(news[:slow])
-            return news[:slow]
 
-        def reversestr(s):
-            left,right = 0,len(s)-1
-            news = list(s)
-            while left<right:
-                news[left],news[right] = news[right],news[left]
-                left+=1; right-=1
-            #return "".join(news)
-            return news
-
-        news = removeextraspace(s)
-        news.append(' ')
-        fast=slow=0
-        #print(news)
-        while fast<len(news):
-            while news[fast]!=' ':
-                fast+=1
-            news[slow:fast] = reversestr(news[slow:fast])
-        #    print(news[slow:fast])
-            fast=slow=fast+1
-        news2 = reversestr(news[:-1])
-        return ''.join(news2)
-```
 Go：
 
 ```go
diff --git a/problems/0202.快乐数.md b/problems/0202.快乐数.md
index 5ac29e86..7fe8cd8d 100644
--- a/problems/0202.快乐数.md
+++ b/problems/0202.快乐数.md
@@ -108,23 +108,14 @@ class Solution {
 ```
 
 Python：
+(版本一)使用集合
 ```python
 class Solution:
-    def isHappy(self, n: int) -> bool:
-        def calculate_happy(num):
-            sum_ = 0
-            
-            # 从个位开始依次取，平方求和
-            while num:
-                sum_ += (num % 10) ** 2
-                num = num // 10
-            return sum_
-
-        # 记录中间结果
+    def isHappy(self, n: int) -> bool:        
         record = set()
 
         while True:
-            n = calculate_happy(n)
+            n = self.get_sum(n)
             if n == 1:
                 return True
             
@@ -134,21 +125,86 @@ class Solution:
             else:
                 record.add(n)
 
-# python的另一种写法 - 通过字符串来计算各位平方和
+    def get_sum(self,n: int) -> int: 
+        new_num = 0
+        while n:
+            n, r = divmod(n, 10)
+            new_num += r ** 2
+        return new_num
+ ```  
+ (版本二)使用集合
+ ```python
+class Solution:
+    def isHappy(self, n: int) -> bool:
+        record = set()
+        while n not in record:
+            record.add(n)
+            new_num = 0
+            n_str = str(n)
+            for i in n_str:
+                new_num+=int(i)**2
+            if new_num==1: return True
+            else: n = new_num
+        return False
+```
+ (版本三)使用数组
+ ```python
 class Solution:
     def isHappy(self, n: int) -> bool:
         record = []
         while n not in record:
             record.append(n)
-            newn = 0
-            nn = str(n)
-            for i in nn:
-                newn+=int(i)**2
-            if newn==1: return True
-            n = newn
+            new_num = 0
+            n_str = str(n)
+            for i in n_str:
+                new_num+=int(i)**2
+            if new_num==1: return True
+            else: n = new_num
         return False
 ```
-
+ (版本四)使用快慢指针
+ ```python
+class Solution:
+    def isHappy(self, n: int) -> bool:        
+        slow = n
+        fast = n
+        while self.get_sum(fast) != 1 and self.get_sum(self.get_sum(fast)):
+            slow = self.get_sum(slow)
+            fast = self.get_sum(self.get_sum(fast))
+            if slow == fast:
+                return False
+        return True
+    def get_sum(self,n: int) -> int: 
+        new_num = 0
+        while n:
+            n, r = divmod(n, 10)
+            new_num += r ** 2
+        return new_num
+```
+ (版本五)使用集合+精简
+ ```python
+class Solution:
+    def isHappy(self, n: int) -> bool:
+        seen = set()
+        while n != 1:
+            n = sum(int(i) ** 2 for i in str(n))
+            if n in seen:
+                return False
+            seen.add(n)
+        return True
+```
+ (版本六)使用数组+精简
+ ```python
+class Solution:
+    def isHappy(self, n: int) -> bool:
+        seen = []
+        while n != 1:
+            n = sum(int(i) ** 2 for i in str(n))
+            if n in seen:
+                return False
+            seen.append(n)
+        return True
+```
 Go：
 ```go
 func isHappy(n int) bool {
diff --git a/problems/0203.移除链表元素.md b/problems/0203.移除链表元素.md
index b52f16ea..6a0de282 100644
--- a/problems/0203.移除链表元素.md
+++ b/problems/0203.移除链表元素.md
@@ -307,21 +307,27 @@ public ListNode removeElements(ListNode head, int val) {
 Python：
 
 ```python
+（版本一）虚拟头节点法
 # Definition for singly-linked list.
 # class ListNode:
 #     def __init__(self, val=0, next=None):
 #         self.val = val
 #         self.next = next
 class Solution:
-    def removeElements(self, head: ListNode, val: int) -> ListNode:
-        dummy_head = ListNode(next=head) #添加一个虚拟节点
-        cur = dummy_head
-        while cur.next:
-            if cur.next.val == val:
-                cur.next = cur.next.next #删除cur.next节点
+    def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:
+        # 创建虚拟头部节点以简化删除过程
+        dummy_head = ListNode(next = head)
+        
+        # 遍历列表并删除值为val的节点
+        current = dummy_head
+        while current.next:
+            if current.next.val == val:
+                current.next = current.next.next
             else:
-                cur = cur.next
+                current = current.next
+        
         return dummy_head.next
+
 ```
 
 Go：
diff --git a/problems/0206.翻转链表.md b/problems/0206.翻转链表.md
index d558e783..0425e182 100644
--- a/problems/0206.翻转链表.md
+++ b/problems/0206.翻转链表.md
@@ -193,9 +193,9 @@ class Solution {
 }
 ```
 
-Python迭代法：
+Python
 ```python
-#双指针
+（版本一）双指针法
 # Definition for singly-linked list.
 # class ListNode:
 #     def __init__(self, val=0, next=None):
@@ -205,7 +205,7 @@ class Solution:
     def reverseList(self, head: ListNode) -> ListNode:
         cur = head   
         pre = None
-        while(cur!=None):
+        while cur:
             temp = cur.next # 保存一下 cur的下一个节点，因为接下来要改变cur->next
             cur.next = pre #反转
             #更新pre、cur指针
@@ -217,6 +217,7 @@ class Solution:
 Python递归法：
 
 ```python
+（版本二）递归法
 # Definition for singly-linked list.
 # class ListNode:
 #     def __init__(self, val=0, next=None):
@@ -224,36 +225,17 @@ Python递归法：
 #         self.next = next
 class Solution:
     def reverseList(self, head: ListNode) -> ListNode:
-        
-        def reverse(pre,cur):
-            if not cur:
-                return pre
-                
-            tmp = cur.next
-            cur.next = pre
-
-            return reverse(cur,tmp)
-        
-        return reverse(None,head)
+        return self.reverse(head, None)
+    def reverse(self, cur: ListNode, pre: ListNode) -> ListNode:
+        if cur == None:
+            return pre
+        temp = cur.next
+        cur.next = pre
+        return self.reverse(temp, cur)
 
 ```
 
-Python递归法从后向前：
 
-```python
-# Definition for singly-linked list.
-# class ListNode:
-#     def __init__(self, val=0, next=None):
-#         self.val = val
-#         self.next = next
-class Solution:
-    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
-        if not head or not head.next: return head
-        p = self.reverseList(head.next)
-        head.next.next = head
-        head.next = None
-        return p
-```
 
 Go：
 
diff --git a/problems/0209.长度最小的子数组.md b/problems/0209.长度最小的子数组.md
index 5a8f91af..d7ae4780 100644
--- a/problems/0209.长度最小的子数组.md
+++ b/problems/0209.长度最小的子数组.md
@@ -173,18 +173,44 @@ class Solution {
 Python：
 
 ```python
+（版本一）滑动窗口法
 class Solution:
     def minSubArrayLen(self, s: int, nums: List[int]) -> int:
-        res = float("inf")   # 定义一个无限大的数
-        Sum = 0     # 滑动窗口数值之和
-        i = 0      # 滑动窗口起始位置
-        for j in range(len(nums)):
-            Sum += nums[j]
-            while Sum >= s:
-                res = min(res, j-i+1)
-                Sum -= nums[i]
-                i += 1
-        return 0 if res == float("inf") else res
+        l = len(nums)
+        left = 0
+        right = 0
+        min_len = float('inf')
+        cur_sum = 0 #当前的累加值
+        
+        while right < l:
+            cur_sum += nums[right]
+            
+            while cur_sum >= s: # 当前累加值大于目标值
+                min_len = min(min_len, right - left + 1)
+                cur_sum -= nums[left]
+                left += 1
+            
+            right += 1
+        
+        return min_len if min_len != float('inf') else 0
+```
+
+```python
+（版本二）暴力法
+class Solution:
+    def minSubArrayLen(self, s: int, nums: List[int]) -> int:
+        l = len(nums)
+        min_len = float('inf')
+        
+        for i in range(l):
+            cur_sum = 0
+            for j in range(i, l):
+                cur_sum += nums[j]
+                if cur_sum >= s:
+                    min_len = min(min_len, j - i + 1)
+                    break
+        
+        return min_len if min_len != float('inf') else 0
 ```
 
 Go：
diff --git a/problems/0222.完全二叉树的节点个数.md b/problems/0222.完全二叉树的节点个数.md
index d89a9bce..795a6f37 100644
--- a/problems/0222.完全二叉树的节点个数.md
+++ b/problems/0222.完全二叉树的节点个数.md
@@ -393,6 +393,20 @@ class Solution: # 利用完全二叉树特性
             return 2**count-1
         return 1+self.countNodes(root.left)+self.countNodes(root.right)  
 ```
+完全二叉树写法3
+```python
+class Solution: # 利用完全二叉树特性
+    def countNodes(self, root: TreeNode) -> int:
+        if not root: return 0
+        count = 0
+        left = root.left; right = root.right
+        while left and right:
+            count+=1
+            left = left.left; right = right.right
+        if not left and not right: # 如果同时到底说明是满二叉树，反之则不是
+            return (2<<count)-1
+        return 1+self.countNodes(root.left)+self.countNodes(root.right)  
+```
 
 ## Go
 
diff --git a/problems/0226.翻转二叉树.md b/problems/0226.翻转二叉树.md
index 16a5be57..63baa409 100644
--- a/problems/0226.翻转二叉树.md
+++ b/problems/0226.翻转二叉树.md
@@ -314,81 +314,158 @@ class Solution {
 
 递归法：前序遍历：
 ```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
 class Solution:
     def invertTree(self, root: TreeNode) -> TreeNode:
         if not root:
             return None
-        root.left, root.right = root.right, root.left #中
-        self.invertTree(root.left) #左
-        self.invertTree(root.right) #右
+        root.left, root.right = root.right, root.left
+        self.invertTree(root.left)
+        self.invertTree(root.right)
         return root
 ```
 
-递归法：后序遍历：
+迭代法：前序遍历：
 ```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
 class Solution:
     def invertTree(self, root: TreeNode) -> TreeNode:
-        if root is None:
+        if not root:
+            return None      
+        stack = [root]        
+        while stack:
+            node = stack.pop()   
+            node.left, node.right = node.right, node.left                   
+            if node.left:
+                stack.append(node.left)
+            if node.right:
+                stack.append(node.right)  
+        return root
+```
+
+
+递归法：中序遍历：
+```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def invertTree(self, root: TreeNode) -> TreeNode:
+        if not root:
+            return None
+        self.invertTree(root.left)
+        root.left, root.right = root.right, root.left
+        self.invertTree(root.left)
+        return root
+```
+
+迭代法：中序遍历：
+```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def invertTree(self, root: TreeNode) -> TreeNode:
+        if not root:
+            return None      
+        stack = [root]        
+        while stack:
+            node = stack.pop()                   
+            if node.left:
+                stack.append(node.left)
+            node.left, node.right = node.right, node.left               
+            if node.left:
+                stack.append(node.left)       
+        return root
+```
+
+
+递归法：后序遍历：
+```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+class Solution:
+    def invertTree(self, root: TreeNode) -> TreeNode:
+        if not root:
             return None
         self.invertTree(root.left)
         self.invertTree(root.right)
         root.left, root.right = root.right, root.left
-        return root 
+        return root
 ```
 
-迭代法：深度优先遍历（前序遍历）：
+迭代法：后序遍历：
 ```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
 class Solution:
     def invertTree(self, root: TreeNode) -> TreeNode:
         if not root:
-            return root
-        st = []
-        st.append(root)
-        while st:
-            node = st.pop()
-            node.left, node.right = node.right, node.left #中
-            if node.right:
-                st.append(node.right) #右
+            return None      
+        stack = [root]        
+        while stack:
+            node = stack.pop()                   
             if node.left:
-                st.append(node.left) #左
+                stack.append(node.left)
+            if node.right:
+                stack.append(node.right)  
+            node.left, node.right = node.right, node.left               
+     
         return root
 ```
 
+
+
+
+
 迭代法：广度优先遍历（层序遍历）：
 ```python
-import collections
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
 class Solution:
     def invertTree(self, root: TreeNode) -> TreeNode:
-        queue = collections.deque() #使用deque()
-        if root:
-            queue.append(root)
+        if not root: 
+            return None
+
+        queue = collections.deque([root])    
         while queue:
-            size = len(queue)
-            for i in range(size):
+            for i in range(len(queue)):
                 node = queue.popleft()
-                node.left, node.right = node.right, node.left #节点处理
-                if node.left:
-                    queue.append(node.left)
-                if node.right:
-                    queue.append(node.right)
-        return root
-```
-迭代法：广度优先遍历（层序遍历），和之前的层序遍历写法一致：
-```python
-class Solution:
-    def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
-        if not root: return root
-        from collections import deque
-        que=deque([root])    
-        while que:
-            size=len(que)
-            for i in range(size):
-                cur=que.popleft()
-                cur.left, cur.right = cur.right, cur.left
-                if cur.left: que.append(cur.left)
-                if cur.right: que.append(cur.right)
+                node.left, node.right = node.right, node.left
+                if node.left: queue.append(node.left)
+                if node.right: queue.append(node.right)
         return root   
+
 ```
+
 ### Go
 
 递归版本的前序遍历
diff --git a/problems/0242.有效的字母异位词.md b/problems/0242.有效的字母异位词.md
index ba802bbd..101dd6f2 100644
--- a/problems/0242.有效的字母异位词.md
+++ b/problems/0242.有效的字母异位词.md
@@ -122,6 +122,7 @@ class Solution {
 ```
 
 Python：
+
 ```python
 class Solution:
     def isAnagram(self, s: str, t: str) -> bool:
@@ -137,7 +138,6 @@ class Solution:
                 return False
         return True
 ```
-
 Python写法二（没有使用数组作为哈希表，只是介绍defaultdict这样一种解题思路）：
 
 ```python
@@ -147,13 +147,11 @@ class Solution:
         
         s_dict = defaultdict(int)
         t_dict = defaultdict(int)
-
         for x in s:
             s_dict[x] += 1
         
         for x in t:
             t_dict[x] += 1
-
         return s_dict == t_dict
 ```
 Python写法三(没有使用数组作为哈希表，只是介绍Counter这种更方便的解题思路)：
@@ -165,7 +163,6 @@ class Solution(object):
         a_count = Counter(s)
         b_count = Counter(t)
         return a_count == b_count
-```
 
 Go：
 
diff --git a/problems/0257.二叉树的所有路径.md b/problems/0257.二叉树的所有路径.md
index 8c542cea..c396f4a0 100644
--- a/problems/0257.二叉树的所有路径.md
+++ b/problems/0257.二叉树的所有路径.md
@@ -468,6 +468,71 @@ class Solution {
 ```
 ---
 ## Python:  
+	
+	
+递归法+回溯（版本一）
+```Python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+import copy
+from typing import List, Optional
+
+class Solution:
+    def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:
+        if not root:
+            return []
+        result = []
+        self.generate_paths(root, [], result)
+        return result
+    
+    def generate_paths(self, node: TreeNode, path: List[int], result: List[str]) -> None:
+        path.append(node.val)
+        if not node.left and not node.right:
+            result.append('->'.join(map(str, path)))
+        if node.left:
+            self.generate_paths(node.left, copy.copy(path), result)
+        if node.right:
+            self.generate_paths(node.right, copy.copy(path), result)
+        path.pop()
+
+
+```
+递归法+回溯（版本二）
+```Python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+import copy
+from typing import List, Optional
+
+class Solution:
+    def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:
+        if not root:
+            return []
+        result = []
+        self.generate_paths(root, [], result)
+        return result
+    
+    def generate_paths(self, node: TreeNode, path: List[int], result: List[str]) -> None:
+        if not node:
+            return
+        path.append(node.val)
+        if not node.left and not node.right:
+            result.append('->'.join(map(str, path)))
+        else:
+            self.generate_paths(node.left, copy.copy(path), result)
+            self.generate_paths(node.right, copy.copy(path), result)
+        path.pop()
+
+```
+	
 递归法+隐形回溯
 ```Python
 # Definition for a binary tree node.
diff --git a/problems/0344.反转字符串.md b/problems/0344.反转字符串.md
index 775cfc58..1c74f9aa 100644
--- a/problems/0344.反转字符串.md
+++ b/problems/0344.反转字符串.md
@@ -174,6 +174,7 @@ class Solution {
 ```
 
 Python：
+（版本一） 双指针
 ```python
 class Solution:
     def reverseString(self, s: List[str]) -> None:
@@ -182,15 +183,70 @@ class Solution:
         """
         left, right = 0, len(s) - 1
         
-        # 该方法已经不需要判断奇偶数，经测试后时间空间复杂度比用 for i in range(right//2)更低
-        # 推荐该写法，更加通俗易懂
+        # 该方法已经不需要判断奇偶数，经测试后时间空间复杂度比用 for i in range(len(s)//2)更低
+        # 因为while每次循环需要进行条件判断，而range函数不需要，直接生成数字，因此时间复杂度更低。推荐使用range
         while left < right:
             s[left], s[right] = s[right], s[left]
             left += 1
             right -= 1
        
 ```
-
+（版本二） 使用栈
+```python
+class Solution:
+    def reverseString(self, s: List[str]) -> None:
+        """
+        Do not return anything, modify s in-place instead.
+        """
+        stack = []
+        for char in s:
+            stack.append(char)
+        for i in range(len(s)):
+            s[i] = stack.pop()
+       
+```
+（版本三） 使用range
+```python
+class Solution:
+    def reverseString(self, s: List[str]) -> None:
+        """
+        Do not return anything, modify s in-place instead.
+        """
+        n = len(s)
+        for i in range(n // 2):
+            s[i], s[n - i - 1] = s[n - i - 1], s[i]
+       
+```
+（版本四） 使用reversed
+```python
+class Solution:
+    def reverseString(self, s: List[str]) -> None:
+        """
+        Do not return anything, modify s in-place instead.
+        """
+        s[:] = reversed(s)
+       
+```
+（版本五） 使用切片
+```python
+class Solution:
+    def reverseString(self, s: List[str]) -> None:
+        """
+        Do not return anything, modify s in-place instead.
+        """
+        s[:] = s[::-1]
+       
+```
+（版本六） 使用列表推导
+```python
+class Solution:
+    def reverseString(self, s: List[str]) -> None:
+        """
+        Do not return anything, modify s in-place instead.
+        """
+        s[:] = [s[i] for i in range(len(s) - 1, -1, -1)]
+       
+```
 Go：
 ```Go
 func reverseString(s []byte) {
diff --git a/problems/0349.两个数组的交集.md b/problems/0349.两个数组的交集.md
index 0da0e30c..c2f6ef46 100644
--- a/problems/0349.两个数组的交集.md
+++ b/problems/0349.两个数组的交集.md
@@ -160,22 +160,29 @@ class Solution {
 ```
 
 Python3：
+（版本一） 使用字典和集合
 ```python
 class Solution:
     def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
-        val_dict = {}
-        ans = []
+    # 使用哈希表存储一个数组中的所有元素
+        table = {}
         for num in nums1:
-            val_dict[num] = 1
-
-        for num in nums2:
-            if num in val_dict.keys() and val_dict[num] == 1:
-                ans.append(num)
-                val_dict[num] = 0
+            table[num] = table.get(num, 0) + 1
         
-        return ans
+        # 使用集合存储结果
+        res = set()
+        for num in nums2:
+            if num in table:
+                res.add(num)
+                del table[num]
+        
+        return list(res)
+```
+（版本二） 使用数组
+
+```python
 	
-class Solution: # 使用数组方法
+class Solution:
     def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
         count1 = [0]*1001
         count2 = [0]*1001
@@ -190,7 +197,14 @@ class Solution: # 使用数组方法
         return result
 	
 ```
+（版本三） 使用集合
 
+```python
+class Solution:
+    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
+        return list(set(nums1) & set(nums2))
+	
+```
 
 Go：
 ```go
diff --git a/problems/0383.赎金信.md b/problems/0383.赎金信.md
index a3f87d4a..c7469d65 100644
--- a/problems/0383.赎金信.md
+++ b/problems/0383.赎金信.md
@@ -146,34 +146,27 @@ class Solution {
 
 ```
 
-Python写法一（使用数组作为哈希表）：
-
+（版本一）使用数组
 ```python
 class Solution:
     def canConstruct(self, ransomNote: str, magazine: str) -> bool:
-
-        arr = [0] * 26
-
-        for x in magazine:    # 记录 magazine里各个字符出现次数
-            arr[ord(x) - ord('a')] += 1
-
-        for x in ransomNote:  # 在arr里对应的字符个数做--操作
-            if arr[ord(x) - ord('a')] == 0:  # 如果没有出现过直接返回
-                return False
-            else:
-                arr[ord(x) - ord('a')] -= 1
-        
-        return True
+        ransom_count = [0] * 26
+        magazine_count = [0] * 26
+        for c in ransomNote:
+            ransom_count[ord(c) - ord('a')] += 1
+        for c in magazine:
+            magazine_count[ord(c) - ord('a')] += 1
+        return all(ransom_count[i] <= magazine_count[i] for i in range(26))
 ```
 
-Python写法二（使用defaultdict）：
+（版本二）使用defaultdict
 
 ```python
+from collections import defaultdict
+
 class Solution:
     def canConstruct(self, ransomNote: str, magazine: str) -> bool:
 
-        from collections import defaultdict
-
         hashmap = defaultdict(int)
 
         for x in magazine:
@@ -181,59 +174,43 @@ class Solution:
 
         for x in ransomNote:
             value = hashmap.get(x)
-            if value is None or value == 0:
+            if not value or not value:
                 return False
             else:
                 hashmap[x] -= 1
 
         return True
 ```
-
-Python写法三：
-
-```python
-class Solution(object):
-    def canConstruct(self, ransomNote, magazine):
-        """
-        :type ransomNote: str
-        :type magazine: str
-        :rtype: bool
-        """
-        
-        # use a dict to store the number of letter occurance in ransomNote
-        hashmap = dict()
-        for s in ransomNote:
-            if s in hashmap:
-                hashmap[s] += 1
-            else:
-                hashmap[s] = 1
-        
-        # check if the letter we need can be found in magazine
-        for l in magazine:
-            if l in hashmap:
-                hashmap[l] -= 1
-        
-        for key in hashmap:
-            if hashmap[key] > 0:
-                return False
-        
-        return True
-```
-
-Python写法四：
+（版本三）使用字典
 
 ```python
 class Solution:
     def canConstruct(self, ransomNote: str, magazine: str) -> bool:
-        c1 = collections.Counter(ransomNote)
-        c2 = collections.Counter(magazine)
-        x = c1 - c2
-        #x只保留值大于0的符号，当c1里面的符号个数小于c2时，不会被保留
-        #所以x只保留下了，magazine不能表达的
-        if(len(x)==0):
-            return True
-        else:
-            return False
+        counts = {}
+        for c in magazine:
+            counts[c] = counts.get(c, 0) + 1
+        for c in ransomNote:
+            if c not in counts or counts[c] == 0:
+                return False
+            counts[c] -= 1
+        return True
+```
+（版本四）使用Counter
+
+```python
+from collections import Counter
+
+class Solution:
+    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
+        return not Counter(ransomNote) - Counter(magazine)
+```
+（版本五）使用count
+
+```python
+class Solution:
+    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
+        return all(ransomNote.count(c) <= magazine.count(c) for c in set(ransomNote))
+       
 ```
 
 Go：
diff --git a/problems/0404.左叶子之和.md b/problems/0404.左叶子之和.md
index cf5441c4..617978b7 100644
--- a/problems/0404.左叶子之和.md
+++ b/problems/0404.左叶子之和.md
@@ -250,19 +250,27 @@ class Solution {
 
 **递归后序遍历**
 ```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
 class Solution:
-    def sumOfLeftLeaves(self, root: TreeNode) -> int:
-        if not root: 
+    def sumOfLeftLeaves(self, root: Optional[TreeNode]) -> int:
+        if not root:
             return 0
         
-        left_left_leaves_sum = self.sumOfLeftLeaves(root.left)  # 左
-        right_left_leaves_sum = self.sumOfLeftLeaves(root.right) # 右
-        
-        cur_left_leaf_val = 0
-        if root.left and not root.left.left and not root.left.right: 
-            cur_left_leaf_val = root.left.val 
+        # 检查根节点的左子节点是否为叶节点
+        if root.left and not root.left.left and not root.left.right:
+            left_val = root.left.val
+        else:
+            left_val = self.sumOfLeftLeaves(root.left)
             
-        return cur_left_leaf_val + left_left_leaves_sum + right_left_leaves_sum # 中
+        # 递归地计算右子树左叶节点的和
+        right_val = self.sumOfLeftLeaves(root.right)
+        
+        return left_val + right_val
 ```
 
 **迭代**
diff --git a/problems/0454.四数相加II.md b/problems/0454.四数相加II.md
index abfc7c23..4a16d4f4 100644
--- a/problems/0454.四数相加II.md
+++ b/problems/0454.四数相加II.md
@@ -118,11 +118,11 @@ class Solution {
 ```
 
 Python：
-
+（版本一） 使用字典
 ```python
 class Solution(object):
     def fourSumCount(self, nums1, nums2, nums3, nums4):
-        # use a dict to store the elements in nums1 and nums2 and their sum
+        # 使用字典存储nums1和nums2中的元素及其和
         hashmap = dict()
         for n1 in nums1:
             for n2 in nums2:
@@ -131,7 +131,7 @@ class Solution(object):
                 else:
                     hashmap[n1+n2] = 1
         
-        # if the -(a+b) exists in nums3 and nums4, we shall add the count
+        # 如果 -(n1+n2) 存在于nums3和nums4, 存入结果
         count = 0
         for n3 in nums3:
             for n4 in nums4:
@@ -142,20 +142,40 @@ class Solution(object):
  
 
 ```
-
+（版本二） 使用字典
 ```python
+class Solution(object):
+    def fourSumCount(self, nums1, nums2, nums3, nums4):
+        # 使用字典存储nums1和nums2中的元素及其和
+        hashmap = dict()
+        for n1 in nums1:
+            for n2 in nums2:
+                hashmap[n1+n2] = hashmap.get(n1+n2, 0) + 1
+        
+        # 如果 -(n1+n2) 存在于nums3和nums4, 存入结果
+        count = 0
+        for n3 in nums3:
+            for n4 in nums4:
+                key = - n3 - n4
+                if key in hashmap:
+                    count += hashmap[key]
+        return count
+ 
+ 
+
+```
+（版本三）使用 defaultdict
+```python
+from collections import defaultdict 
 class Solution:
     def fourSumCount(self, nums1: list, nums2: list, nums3: list, nums4: list) -> int:
-        from collections import defaultdict # You may use normal dict instead.
         rec, cnt = defaultdict(lambda : 0), 0
-        # To store the summary of all the possible combinations of nums1 & nums2, together with their frequencies.
         for i in nums1:
             for j in nums2:
                 rec[i+j] += 1
-        # To add up the frequencies if the corresponding value occurs in the dictionary
         for i in nums3:
             for j in nums4:
-                cnt += rec.get(-(i+j), 0) # No matched key, return 0.
+                cnt += rec.get(-(i+j), 0) 
         return cnt
 ```
 
diff --git a/problems/0459.重复的子字符串.md b/problems/0459.重复的子字符串.md
index 98c02a25..5d56ad18 100644
--- a/problems/0459.重复的子字符串.md
+++ b/problems/0459.重复的子字符串.md
@@ -264,8 +264,7 @@ class Solution {
 
 Python：
 
-这里使用了前缀表统一减一的实现方式
-
+（版本一） 前缀表 减一
 ```python
 class Solution:
     def repeatedSubstringPattern(self, s: str) -> bool:  
@@ -289,7 +288,7 @@ class Solution:
         return nxt
 ```
 
-前缀表（不减一）的代码实现
+（版本二） 前缀表 不减一
 
 ```python
 class Solution:
@@ -314,6 +313,40 @@ class Solution:
         return nxt
 ```
 
+
+（版本三） 使用 find
+
+```python
+class Solution:
+    def repeatedSubstringPattern(self, s: str) -> bool:
+        n = len(s)
+        if n <= 1:
+            return False
+        ss = s[1:] + s[:-1] 
+        print(ss.find(s))              
+        return ss.find(s) != -1
+```
+
+（版本四） 暴力法
+
+```python
+class Solution:
+    def repeatedSubstringPattern(self, s: str) -> bool:
+        n = len(s)
+        if n <= 1:
+            return False
+        
+        substr = ""
+        for i in range(1, n//2 + 1):
+            if n % i == 0:
+                substr = s[:i]
+                if substr * (n//i) == s:
+                    return True
+                
+        return False
+```
+
+
 Go：
 
 这里使用了前缀表统一减一的实现方式
diff --git a/problems/0513.找树左下角的值.md b/problems/0513.找树左下角的值.md
index 5ff25484..26768c74 100644
--- a/problems/0513.找树左下角的值.md
+++ b/problems/0513.找树左下角的值.md
@@ -271,52 +271,80 @@ class Solution {
 
 
 ### Python 
-
-递归:
+（版本一）递归法 + 回溯
 ```python
 class Solution:
     def findBottomLeftValue(self, root: TreeNode) -> int:
-        max_depth = -float("INF")
-        leftmost_val = 0
+        self.max_depth = float('-inf')
+        self.result = None
+        self.traversal(root, 0)
+        return self.result
+    
+    def traversal(self, node, depth):
+        if not node.left and not node.right:
+            if depth > self.max_depth:
+                self.max_depth = depth
+                self.result = node.val
+            return
+        
+        if node.left:
+            depth += 1
+            self.traversal(node.left, depth)
+            depth -= 1
+        if node.right:
+            depth += 1
+            self.traversal(node.right, depth)
+            depth -= 1
 
-        def __traverse(root, cur_depth): 
-            nonlocal max_depth, leftmost_val
-            if not root.left and not root.right: 
-                if cur_depth > max_depth: 
-                    max_depth = cur_depth
-                    leftmost_val = root.val  
-            if root.left: 
-                cur_depth += 1
-                __traverse(root.left, cur_depth)
-                cur_depth -= 1
-            if root.right: 
-                cur_depth += 1
-                __traverse(root.right, cur_depth)
-                cur_depth -= 1
-
-        __traverse(root, 0)
-        return leftmost_val
 ```
 
-迭代 - 层序遍历:
+（版本二）递归法+精简
 ```python
 class Solution:
     def findBottomLeftValue(self, root: TreeNode) -> int:
-        queue = deque()
-        if root: 
-            queue.append(root)
-        result = 0
-        while queue: 
-            q_len = len(queue)
-            for i in range(q_len): 
-                if i == 0: 
-                    result = queue[i].val 
-                cur = queue.popleft()
-                if cur.left: 
-                    queue.append(cur.left)
-                if cur.right: 
-                    queue.append(cur.right)
-        return result
+        self.max_depth = float('-inf')
+        self.result = None
+        self.traversal(root, 0)
+        return self.result
+    
+    def traversal(self, node, depth):
+        if not node.left and not node.right:
+            if depth > self.max_depth:
+                self.max_depth = depth
+                self.result = node.val
+            return
+        
+        if node.left:
+            self.traversal(node.left, depth+1)
+        if node.right:
+            self.traversal(node.right, depth+1)
+```
+
+(版本三） 迭代法
+```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+from collections import deque
+class Solution:
+    def findBottomLeftValue(self, root: Optional[TreeNode]) -> int:
+        queue = deque([root])
+        while queue:
+            size = len(queue)
+            leftmost = queue[0].val
+            for i in range(size):
+                node = queue.popleft()
+                if node.left:
+                    queue.append(node.left)
+                if node.right:
+                    queue.append(node.right)
+            if not queue:
+                return leftmost
+
+
 ```
 
 ### Go
diff --git a/problems/0647.回文子串.md b/problems/0647.回文子串.md
index 90e6da9f..521c8c26 100644
--- a/problems/0647.回文子串.md
+++ b/problems/0647.回文子串.md
@@ -267,6 +267,27 @@ class Solution {
         return ans;
     }
 }
+
+```
+
+动态规划：简洁版
+```java
+class Solution {
+    public int countSubstrings(String s) {
+        boolean[][] dp = new boolean[s.length()][s.length()];
+        
+        int res = 0;
+        for (int i = s.length() - 1; i >= 0; i--) {
+            for (int j = i; j < s.length(); j++) {
+                if (s.charAt(i) == s.charAt(j) && (j - i <= 1 || dp[i + 1][j - 1])) {
+                    res++;
+                    dp[i][j] = true;
+                }
+            }
+        }
+        return res;
+    }
+}
 ```
 
 中心扩散法：
diff --git a/problems/0707.设计链表.md b/problems/0707.设计链表.md
index 5c78a12a..aa04d0e1 100644
--- a/problems/0707.设计链表.md
+++ b/problems/0707.设计链表.md
@@ -485,177 +485,176 @@ class MyLinkedList {
 
 Python：
 ```python
-# 单链表
-class Node(object):
-    def __init__(self, x=0):
-        self.val = x
-        self.next = None
-
-class MyLinkedList(object):
-
-    def __init__(self):
-        self.head = Node()
-        self.size = 0 # 设置一个链表长度的属性，便于后续操作，注意每次增和删的时候都要更新
-
-    def get(self, index):
-        """
-        :type index: int
-        :rtype: int
-        """
-        if index < 0 or index >= self.size:
-            return -1
-        cur = self.head.next
-        while(index):
-            cur = cur.next
-            index -= 1
-        return cur.val
-
-    def addAtHead(self, val):
-        """
-        :type val: int
-        :rtype: None
-        """
-        new_node = Node(val)
-        new_node.next = self.head.next
-        self.head.next = new_node
-        self.size += 1
-
-    def addAtTail(self, val):
-        """
-        :type val: int
-        :rtype: None
-        """
-        new_node = Node(val)
-        cur = self.head
-        while(cur.next):
-            cur = cur.next
-        cur.next = new_node
-        self.size += 1
-
-    def addAtIndex(self, index, val):
-        """
-        :type index: int
-        :type val: int
-        :rtype: None
-        """
-        if index < 0:
-            self.addAtHead(val)
-            return
-        elif index == self.size:
-            self.addAtTail(val)
-            return
-        elif index > self.size:
-            return
-
-        node = Node(val)
-        pre = self.head
-        while(index):
-            pre = pre.next
-            index -= 1
-        node.next = pre.next
-        pre.next = node
-        self.size += 1
-        
-    def deleteAtIndex(self, index):
-        """
-        :type index: int
-        :rtype: None
-        """
-        if index < 0 or index >= self.size:
-            return
-        pre = self.head
-        while(index):
-            pre = pre.next
-            index -= 1
-        pre.next = pre.next.next
-        self.size -= 1
-	
-# 双链表
-# 相对于单链表, Node新增了prev属性
-class Node:
-    
-    def __init__(self, val):
+（版本一）单链表法
+class ListNode:
+    def __init__(self, val=0, next=None):
         self.val = val
-        self.prev = None
-        self.next = None
-
-
+        self.next = next
+        
 class MyLinkedList:
-
     def __init__(self):
-        self._head, self._tail = Node(0), Node(0)  # 虚拟节点
-        self._head.next, self._tail.prev = self._tail, self._head
-        self._count = 0  # 添加的节点数
-
-    def _get_node(self, index: int) -> Node:
-        # 当index小于_count//2时, 使用_head查找更快, 反之_tail更快
-        if index >= self._count // 2:
-            # 使用prev往前找
-            node = self._tail
-            for _ in range(self._count - index):
-                node = node.prev
-        else:
-            # 使用next往后找
-            node = self._head   
-            for _ in range(index + 1):
-                node = node.next
-        return node
+        self.dummy_head = ListNode()
+        self.size = 0
 
     def get(self, index: int) -> int:
-        """
-        Get the value of the index-th node in the linked list. If the index is invalid, return -1.
-        """
-        if 0 <= index < self._count:
-            node = self._get_node(index)
-            return node.val
-        else:
+        if index < 0 or index >= self.size:
             return -1
+        
+        current = self.dummy_head.next
+        for i in range(index):
+            current = current.next
+            
+        return current.val
 
     def addAtHead(self, val: int) -> None:
-        """
-        Add a node of value val before the first element of the linked list. After the insertion, the new node will be the first node of the linked list.
-        """
-        self._update(self._head, self._head.next, val)
+        self.dummy_head.next = ListNode(val, self.dummy_head.next)
+        self.size += 1
 
     def addAtTail(self, val: int) -> None:
-        """
-        Append a node of value val to the last element of the linked list.
-        """
-        self._update(self._tail.prev, self._tail, val)
+        current = self.dummy_head
+        while current.next:
+            current = current.next
+        current.next = ListNode(val)
+        self.size += 1
 
     def addAtIndex(self, index: int, val: int) -> None:
-        """
-        Add a node of value val before the index-th node in the linked list. If index equals to the length of linked list, the node will be appended to the end of linked list. If index is greater than the length, the node will not be inserted.
-        """
-        if index < 0:
-            index = 0
-        elif index > self._count:
+        if index < 0 or index > self.size:
             return
-        node = self._get_node(index)
-        self._update(node.prev, node, val)
-
-    def _update(self, prev: Node, next: Node, val: int) -> None:
-        """
-            更新节点
-            :param prev: 相对于更新的前一个节点
-            :param next: 相对于更新的后一个节点
-            :param val:  要添加的节点值
-        """
-        # 计数累加
-        self._count += 1
-        node = Node(val)
-        prev.next, next.prev = node, node
-        node.prev, node.next = prev, next
+        
+        current = self.dummy_head
+        for i in range(index):
+            current = current.next
+        current.next = ListNode(val, current.next)
+        self.size += 1
 
     def deleteAtIndex(self, index: int) -> None:
-        """
-        Delete the index-th node in the linked list, if the index is valid.
-        """
-        if 0 <= index < self._count:
-            node = self._get_node(index)
-            # 计数-1
-            self._count -= 1
-            node.prev.next, node.next.prev = node.next, node.prev
+        if index < 0 or index >= self.size:
+            return
+        
+        current = self.dummy_head
+        for i in range(index):
+            current = current.next
+        current.next = current.next.next
+        self.size -= 1
+
+
+# Your MyLinkedList object will be instantiated and called as such:
+# obj = MyLinkedList()
+# param_1 = obj.get(index)
+# obj.addAtHead(val)
+# obj.addAtTail(val)
+# obj.addAtIndex(index,val)
+# obj.deleteAtIndex(index)
+```
+
+
+```python
+（版本二）双链表法
+class ListNode:
+    def __init__(self, val=0, prev=None, next=None):
+        self.val = val
+        self.prev = prev
+        self.next = next
+
+class MyLinkedList:
+    def __init__(self):
+        self.head = None
+        self.tail = None
+        self.size = 0
+
+    def get(self, index: int) -> int:
+        if index < 0 or index >= self.size:
+            return -1
+        
+        if index < self.size // 2:
+            current = self.head
+            for i in range(index):
+                current = current.next
+        else:
+            current = self.tail
+            for i in range(self.size - index - 1):
+                current = current.prev
+                
+        return current.val
+
+    def addAtHead(self, val: int) -> None:
+        new_node = ListNode(val, None, self.head)
+        if self.head:
+            self.head.prev = new_node
+        else:
+            self.tail = new_node
+        self.head = new_node
+        self.size += 1
+
+    def addAtTail(self, val: int) -> None:
+        new_node = ListNode(val, self.tail, None)
+        if self.tail:
+            self.tail.next = new_node
+        else:
+            self.head = new_node
+        self.tail = new_node
+        self.size += 1
+
+    def addAtIndex(self, index: int, val: int) -> None:
+        if index < 0 or index > self.size:
+            return
+        
+        if index == 0:
+            self.addAtHead(val)
+        elif index == self.size:
+            self.addAtTail(val)
+        else:
+            if index < self.size // 2:
+                current = self.head
+                for i in range(index - 1):
+                    current = current.next
+            else:
+                current = self.tail
+                for i in range(self.size - index):
+                    current = current.prev
+            new_node = ListNode(val, current, current.next)
+            current.next.prev = new_node
+            current.next = new_node
+            self.size += 1
+
+    def deleteAtIndex(self, index: int) -> None:
+        if index < 0 or index >= self.size:
+            return
+        
+        if index == 0:
+            self.head = self.head.next
+            if self.head:
+                self.head.prev = None
+            else:
+                self.tail = None
+        elif index == self.size - 1:
+            self.tail = self.tail.prev
+            if self.tail:
+                self.tail.next = None
+            else:
+                self.head = None
+        else:
+            if index < self.size // 2:
+                current = self.head
+                for i in range(index):
+                    current = current.next
+            else:
+                current = self.tail
+                for i in range(self.size - index - 1):
+                    current = current.prev
+            current.prev.next = current.next
+            current.next.prev = current.prev
+        self.size -= 1
+
+
+
+# Your MyLinkedList object will be instantiated and called as such:
+# obj = MyLinkedList()
+# param_1 = obj.get(index)
+# obj.addAtHead(val)
+# obj.addAtTail(val)
+# obj.addAtIndex(index,val)
+# obj.deleteAtIndex(index)
 ```
 
 Go：
diff --git a/problems/0977.有序数组的平方.md b/problems/0977.有序数组的平方.md
index 57f8de02..4fbdd1cd 100644
--- a/problems/0977.有序数组的平方.md
+++ b/problems/0977.有序数组的平方.md
@@ -140,22 +140,37 @@ class Solution {
 
 Python：
 ```Python
+（版本一）双指针法
 class Solution:
     def sortedSquares(self, nums: List[int]) -> List[int]:
-        n = len(nums)
-        i,j,k = 0,n - 1,n - 1
-        ans = [-1] * n
-        while i <= j:
-            lm = nums[i] ** 2
-            rm = nums[j] ** 2
-            if lm > rm:
-                ans[k] = lm
-                i += 1
+        l, r, i = 0, len(nums)-1, len(nums)-1
+        res = [float('inf')] * len(nums) # 需要提前定义列表，存放结果
+        while l <= r:
+            if nums[l] ** 2 < nums[r] ** 2: # 左右边界进行对比，找出最大值
+                res[i] = nums[r] ** 2
+                r -= 1 # 右指针往左移动
             else:
-                ans[k] = rm
-                j -= 1
-            k -= 1
-        return ans
+                res[i] = nums[l] ** 2
+                l += 1 # 左指针往右移动
+            i -= 1 # 存放结果的指针需要往前平移一位
+        return res
+```
+
+```Python
+（版本二）暴力排序法
+class Solution:
+    def sortedSquares(self, nums: List[int]) -> List[int]:
+        for i in range(len(nums)):
+            nums[i] *= nums[i]
+        nums.sort()
+        return nums
+```
+
+```Python
+（版本三）暴力排序法+列表推导法
+class Solution:
+    def sortedSquares(self, nums: List[int]) -> List[int]:
+        return sorted(x*x for x in nums)
 ```
 
 Go：
diff --git a/problems/二叉树的迭代遍历.md b/problems/二叉树的迭代遍历.md
index 35cf4077..8b241465 100644
--- a/problems/二叉树的迭代遍历.md
+++ b/problems/二叉树的迭代遍历.md
@@ -258,7 +258,9 @@ class Solution:
             if node.left:
                 stack.append(node.left)
         return result
-        
+```
+```python
+
 # 中序遍历-迭代-LC94_二叉树的中序遍历
 class Solution:
     def inorderTraversal(self, root: TreeNode) -> List[int]:
@@ -279,7 +281,9 @@ class Solution:
                 # 取栈顶元素右结点
                 cur = cur.right	
         return result
-        
+ ```
+ ```python
+
 # 后序遍历-迭代-LC145_二叉树的后序遍历
 class Solution:
     def postorderTraversal(self, root: TreeNode) -> List[int]:
diff --git a/problems/二叉树的递归遍历.md b/problems/二叉树的递归遍历.md
index 5a9a670a..92f342f0 100644
--- a/problems/二叉树的递归遍历.md
+++ b/problems/二叉树的递归遍历.md
@@ -174,50 +174,49 @@ class Solution {
 Python：
 ```python
 # 前序遍历-递归-LC144_二叉树的前序遍历
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, val=0, left=None, right=None):
+#         self.val = val
+#         self.left = left
+#         self.right = right
+
 class Solution:
     def preorderTraversal(self, root: TreeNode) -> List[int]:
-        # 保存结果
-        result = []
-        
-        def traversal(root: TreeNode):
-            if root == None:
-                return
-            result.append(root.val) # 前序
-            traversal(root.left)    # 左
-            traversal(root.right)   # 右
+        if not root:
+            return []
 
-        traversal(root)
-        return result
+        left = self.preorderTraversal(root.left)
+        right = self.preorderTraversal(root.right)
 
+        return  [root.val] + left +  right
+
+```
+```python
 # 中序遍历-递归-LC94_二叉树的中序遍历
 class Solution:
     def inorderTraversal(self, root: TreeNode) -> List[int]:
-        result = []
+        if root is None:
+            return []
 
-        def traversal(root: TreeNode):
-            if root == None:
-                return
-            traversal(root.left)    # 左
-            result.append(root.val) # 中序
-            traversal(root.right)   # 右
+        left = self.inorderTraversal(root.left)
+        right = self.inorderTraversal(root.right)
+
+        return left + [root.val] + right
+```
+```python
 
-        traversal(root)
-        return result
 
 # 后序遍历-递归-LC145_二叉树的后序遍历
 class Solution:
     def postorderTraversal(self, root: TreeNode) -> List[int]:
-        result = []
+        if not root:
+            return []
 
-        def traversal(root: TreeNode):
-            if root == None:
-                return
-            traversal(root.left)    # 左
-            traversal(root.right)   # 右
-            result.append(root.val) # 后序
+        left = self.postorderTraversal(root.left)
+        right = self.postorderTraversal(root.right)
 
-        traversal(root)
-        return result
+        return left + right + [root.val]
 ```
 
 Go：
diff --git a/problems/前序/刷力扣用不用库函数.md b/problems/前序/刷力扣用不用库函数.md
index ae0940bf..7d0e3475 100644
--- a/problems/前序/刷力扣用不用库函数.md
+++ b/problems/前序/刷力扣用不用库函数.md
@@ -24,5 +24,5 @@
 
 例如for循环里套一个字符串的insert，erase之类的操作，你说时间复杂度是多少呢，很明显是O(n^2)的时间复杂度了。
 
-在刷题的时候本着我说的标准来使用库函数，详细对大家回有所帮助！
+在刷题的时候本着我说的标准来使用库函数，相信对大家会有所帮助！
 
diff --git a/problems/剑指Offer05.替换空格.md b/problems/剑指Offer05.替换空格.md
index 2e1ee1de..dbad781e 100644
--- a/problems/剑指Offer05.替换空格.md
+++ b/problems/剑指Offer05.替换空格.md
@@ -266,6 +266,8 @@ func replaceSpace(s string) string {
 
 
 python：
+#### 因为字符串是不可变类型，所以操作字符串需要将其转换为列表，因此空间复杂度不可能为O(1)
+（版本一）转换成列表，并且添加相匹配的空间，然后进行填充
 ```python
 class Solution:
     def replaceSpace(self, s: str) -> str:
@@ -290,14 +292,22 @@ class Solution:
         return ''.join(res)
             
 ```
-
+（版本二）添加空列表，添加匹配的结果
+```python
+class Solution:
+    def replaceSpace(self, s: str) -> str:
+        res = []
+        for i in range(len(s)):
+            if s[i] == ' ':
+                res.append('%20')
+            else:
+                res.append(s[i])
+        return ''.join(res)
+```
+（版本三）使用切片
 ```python
 class Solution:
     def replaceSpace(self, s: str) -> str:
-        # method 1 - Very rude
-        return "%20".join(s.split(" "))
-
-        # method 2 - Reverse the s when counting in for loop, then update from the end.
         n = len(s)
         for e, i in enumerate(s[::-1]):
             print(i, e)
@@ -306,7 +316,18 @@ class Solution:
             print("")
         return s
 ```
-
+（版本四）使用join + split
+```python
+class Solution:
+    def replaceSpace(self, s: str) -> str:
+        return "%20".join(s.split(" "))
+```
+（版本五）使用replace
+```python
+class Solution:
+    def replaceSpace(self, s: str) -> str:
+        return s.replace(' ', '%20')
+```
 javaScript:
 
 ```js
diff --git a/problems/剑指Offer58-II.左旋转字符串.md b/problems/剑指Offer58-II.左旋转字符串.md
index f368514f..6cd88456 100644
--- a/problems/剑指Offer58-II.左旋转字符串.md
+++ b/problems/剑指Offer58-II.左旋转字符串.md
@@ -142,15 +142,16 @@ class Solution {
 ```
 
 python: 
+（版本一）使用切片
 
 ```python
-# 方法一：可以使用切片方法
 class Solution:
     def reverseLeftWords(self, s: str, n: int) -> str:
-        return s[n:] + s[0:n]
+        return s[n:] + s[:n]
 ```
+（版本二）使用reversed + join
+
 ```python    
-# 方法二：也可以使用上文描述的方法，有些面试中不允许使用切片，那就使用上文作者提到的方法
 class Solution:
     def reverseLeftWords(self, s: str, n: int) -> str:
         s = list(s)
@@ -161,32 +162,29 @@ class Solution:
         return "".join(s)
 
 ```
+（版本三）自定义reversed函数
 
 ```python
-# 方法三：如果连reversed也不让使用，那么自己手写一个
 class Solution:
-    def reverseLeftWords(self, s: str, n: int) -> str:
-        def reverse_sub(lst, left, right):
-            while left < right:
-                lst[left], lst[right] = lst[right], lst[left]
-                left += 1
-                right -= 1
+    def reverseLeftWords(self, s: str, n: int) -> str:       
+        s_list = list(s)
         
-        res = list(s)
-        end = len(res) - 1
-        reverse_sub(res, 0, n - 1)
-        reverse_sub(res, n, end)
-        reverse_sub(res, 0, end)
-        return ''.join(res)
+        self.reverse(s_list, 0, n - 1)        
+        self.reverse(s_list, n, len(s_list) - 1)
+        self.reverse(s_list, 0, len(s_list) - 1)
 
-# 同方法二
-# 时间复杂度：O(n)
-# 空间复杂度：O(n)，python的string为不可变，需要开辟同样大小的list空间来修改
+        return ''.join(s_list)
+        
+    def reverse(self, s, start, end):
+        while start < end:
+            s[start], s[end] = s[end], s[start]
+            start += 1
+            end -= 1   
 
 ```
+（版本四）使用 模 +下标
 
 ```python 3
-#方法四：考虑不能用切片的情况下，利用模+下标实现
 class Solution:
     def reverseLeftWords(self, s: str, n: int) -> str:
         new_s = ''
@@ -196,17 +194,21 @@ class Solution:
         return new_s
 
 ```
+（版本五）使用 模 + 切片
 
 ```python 3 
-# 方法五：另类的切片方法
 class Solution:
     def reverseLeftWords(self, s: str, n: int) -> str:
-        n = len(s)
-        s = s + s 
-        return s[k : n+k]
+        l = len(s)
+        # 复制输入字符串与它自己连接
+        s = s + s
+        
+        # 计算旋转字符串的起始索引
+        k = n % (l * 2)
+        
+        # 从连接的字符串中提取旋转后的字符串并返回
+        return s[k : k + l]
 
-# 时间复杂度：O(n)
-# 空间复杂度：O(n)
 ```
 
 Go：
diff --git a/problems/面试题02.07.链表相交.md b/problems/面试题02.07.链表相交.md
index 9e0c29f5..75e03116 100644
--- a/problems/面试题02.07.链表相交.md
+++ b/problems/面试题02.07.链表相交.md
@@ -150,9 +150,13 @@ public class Solution {
 }
 ```
 
+
 Python：
 
 ```python
+
+（版本一）求长度，同时出发
+
 class Solution:
     def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
         lenA, lenB = 0, 0
@@ -178,8 +182,105 @@ class Solution:
                 curB = curB.next
         return None 
 ```
+```python
+（版本二）求长度，同时出发 （代码复用）
+class Solution:
+    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
+        lenA = self.getLength(headA)
+        lenB = self.getLength(headB)
+        
+        # 通过移动较长的链表，使两链表长度相等
+        if lenA > lenB:
+            headA = self.moveForward(headA, lenA - lenB)
+        else:
+            headB = self.moveForward(headB, lenB - lenA)
+        
+        # 将两个头向前移动，直到它们相交
+        while headA and headB:
+            if headA == headB:
+                return headA
+            headA = headA.next
+            headB = headB.next
+        
+        return None
+    
+    def getLength(self, head: ListNode) -> int:
+        length = 0
+        while head:
+            length += 1
+            head = head.next
+        return length
+    
+    def moveForward(self, head: ListNode, steps: int) -> ListNode:
+        while steps > 0:
+            head = head.next
+            steps -= 1
+        return head
+```
+```python
+（版本三）求长度，同时出发 （代码复用 + 精简）
+class Solution:
+    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
+        dis = self.getLength(headA) - self.getLength(headB)
+        
+        # 通过移动较长的链表，使两链表长度相等
+        if dis > 0:
+            headA = self.moveForward(headA, dis)
+        else:
+            headB = self.moveForward(headB, abs(dis))
+        
+        # 将两个头向前移动，直到它们相交
+        while headA and headB:
+            if headA == headB:
+                return headA
+            headA = headA.next
+            headB = headB.next
+        
+        return None
+    
+    def getLength(self, head: ListNode) -> int:
+        length = 0
+        while head:
+            length += 1
+            head = head.next
+        return length
+    
+    def moveForward(self, head: ListNode, steps: int) -> ListNode:
+        while steps > 0:
+            head = head.next
+            steps -= 1
+        return head
+```
+```python
+（版本四）等比例法
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
 
-Go：
+
+class Solution:
+    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
+        # 处理边缘情况
+        if not headA or not headB:
+            return None
+        
+        # 在每个链表的头部初始化两个指针
+        pointerA = headA
+        pointerB = headB
+        
+        # 遍历两个链表直到指针相交
+        while pointerA != pointerB:
+            # 将指针向前移动一个节点
+            pointerA = pointerA.next if pointerA else headB
+            pointerB = pointerB.next if pointerB else headA
+        
+        # 如果相交，指针将位于交点节点，如果没有交点，值为None
+        return pointerA
+```
+
+Go:
 
 ```go
 func getIntersectionNode(headA, headB *ListNode) *ListNode {