Merge pull request #2075 from jianghongcheng/master

跟新python代码，已经检查

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()
+        # 创建一个虚拟节点，并将其下一个指针设置为链表的头部
-        head_dummy.next = head
+        dummy_head = ListNode(0, head)
         
-        slow, fast = head_dummy, head_dummy
+        # 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
-        while(n>=0): #fast先往前走n+1步
+        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
+        # 通过更新第 (n-1) 个节点的 next 指针删除第 n 个节点
-        return head_dummy.next
+        slow.next = slow.next.next
+        
+        return dummy_head.next
+
 ```
 Go:
 ```Go

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)
+        dummy_head = ListNode(next=head)
-        pre = res
+        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
+        # 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 pre.next and pre.next.next:
+        while current.next and current.next.next:
-            cur = pre.next
+            temp = current.next # 防止节点修改
-            post = pre.next.next
+            temp1 = current.next.next.next
             
-            # pre，cur，post对应最左，中间的，最右边的节点
+            current.next = current.next.next
-            cur.next = post.next
+            current.next.next = temp
-            post.next = cur
+            temp.next = temp1
-            pre.next = post
+            current = current.next.next
+        return dummy_head.next
 
-            pre = pre.next.next
-        return res.next
 ```
 
 Go：

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：

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]
+        queue = collections.deque([root.left, root.right])
-        while que:
+        
-            this_level_length = len(que)
+        while queue:
-            for i in range(this_level_length // 2):
+            level_size = len(queue)
-                # 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]):
+            if level_size % 2 != 0:
-                    return False
+                return False
-                # 要么两个都不是None
+            
-                if que[i] and que[i].val != que[this_level_length - 1 - i].val:
+            level_vals = []
-                    return False
+            for i in range(level_size):
-            for i in range(this_level_length):
+                node = queue.popleft()
-                if not que[i]: continue
+                if node:
-                que.append(que[i].left)
+                    level_vals.append(node.val)
-                que.append(que[i].right)
+                    queue.append(node.left)
-            que = que[this_level_length:]
+                    queue.append(node.right)
+                else:
+                    level_vals.append(None)
+                    
+            if level_vals != level_vals[::-1]:
+                return False
+            
         return True
 ```
 

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: Optional[TreeNode]) -> List[List[int]]:
-
-    def levelOrder(self, root: TreeNode) -> List[List[int]]:
-        results = []
         if not root:
-            return results
+            return []
-
+        queue = collections.deque([root])
-        from collections import deque
+        result = []
-        que = deque([root])
+        while queue:
-
+            level = []
-        while que:
+            for _ in range(len(queue)):
-            size = len(que)
+                cur = queue.popleft()
-            result = []
+                level.append(cur.val)
-            for _ in range(size):
-                cur = que.popleft()
-                result.append(cur.val)
                 if cur.left:
-                    que.append(cur.left)
+                    queue.append(cur.left)
                 if cur.right:
-                    que.append(cur.right)
+                    queue.append(cur.right)
-            results.append(result)
+            result.append(level)
-
+        return result
-        return results
 ```
-
 ```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]]:
+    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
-        res = []
+        levels = []
-        def helper(root, depth):
+        self.helper(root, 0, levels)
-            if not root: return []
+        return levels
-            if len(res) == depth: res.append([]) # start the current depth
+    
-            res[depth].append(root.val) # fulfil the current depth
+    def helper(self, node, level, levels):
-            if  root.left: helper(root.left, depth + 1) # process child nodes for the next depth
+        if not node:
-            if  root.right: helper(root.right, depth + 1)
+            return
-        helper(root, 0)
+        if len(levels) == level:
-        return res
+            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
+            return []
-
+        queue = collections.deque([root])
-        from collections import deque
+        result = []
-        que = deque([root])
+        while queue:
-
+            level = []
-        while que:
+            for _ in range(len(queue)):
-            result = []
+                cur = queue.popleft()
-            for _ in range(len(que)):
+                level.append(cur.val)
-                cur = que.popleft()
-                result.append(cur.val)
                 if cur.left:
-                    que.append(cur.left)
+                    queue.append(cur.left)
                 if cur.right:
-                    que.append(cur.right)
+                    queue.append(cur.right)
-            results.append(result)
+            result.append(level)
-
+        return result[::-1]
-        results.reverse()
-        return results
 ```
 
 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
+        queue = collections.deque([root])
-        # 'from collections import deque' 导入
+        right_view = []
-        # 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])
+        while queue:
-        out_list = []
+            level_size = len(queue)
             
-        while quene:
+            for i in range(level_size):
-            # 每次都取最后一个node就可以了
+                node = queue.popleft()
-            node = quene[-1]
+                
-            out_list.append(node.val)
+                if i == level_size - 1:
+                    right_view.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()
                 if node.left:
-                    quene.append(node.left)
+                    queue.append(node.left)
                 if node.right:
-                    quene.append(node.right)
+                    queue.append(node.right)
         
-        return out_list
+        return right_view
-
-# 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
 ```
 
 
@@ -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
+        queue = collections.deque([root])
-        que = deque([root])
+        averages = []
         
-        while que:
+        while queue:
-            size = len(que)
+            size = len(queue)
-            sum_ = 0
+            level_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
+            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
+        result = []
-        que = deque([root])
+        queue = collections.deque([root])
 
-        while que:
+        while queue:
-            result = []
+            level_size = len(queue)
-            for _ in range(len(que)):
+            level = []
-                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)
 
-        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)
+            level_size = len(queue)
-            in_list = []
+            max_val = float('-inf')
-            for _ in range(length):
+
-                curnode = queue.pop(0)
+            for _ in range(level_size):
-                in_list.append(curnode.val)
+                node = queue.popleft()
-                if curnode.left: queue.append(curnode.left)
+                max_val = max(max_val, node.val)
-                if curnode.right: queue.append(curnode.right)
+
-            out_list.append(max(in_list))
+                if node.left:
-        return out_list
+                    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
+            return root
-        queue = [root]
+        
+        queue = collections.deque([root])
+        
         while queue:
-            n = len(queue)
+            level_size = len(queue)
-            for i in range(n):
+            prev = None
-                node = queue.pop(0)
+            
+            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
+            return root
-        queue = [root]
+        
-        while queue:  # 遍历每一层
+        queue = collections.deque([root])
-            length = len(queue)
+        
-            tail = None # 每一层维护一个尾节点
+        while queue:
-            for i in range(length): # 遍历当前层
+            level_size = len(queue)
-                curnode = queue.pop(0)
+            prev = None
-                if tail:
+            
-                    tail.next = curnode # 让尾节点指向当前节点
+            for i in range(level_size):
-                tail = curnode # 让当前节点成为尾节点
+                node = queue.popleft()
-                if curnode.left : queue.append(curnode.left)
+                
-                if curnode.right: queue.append(curnode.right)
+                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_:
+        queue = collections.deque([root])
-            length = len(queue_)
+        
-            for i in range(length):
+        while queue:
-                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)
             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])
         
-        #根节点的深度为1
+        while queue:
-        queue_ = [(root,1)]
+            depth += 1 
-        while queue_:
+            for _ in range(len(queue)):
-            cur, depth = queue_.pop(0)
+                node = queue.popleft()
                 
-            if cur.left == None and cur.right == None:
+                if not node.left and not node.right:
-                return depth
+                    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
+                if node.left:
+                    queue.append(node.left)
+                    
+                if node.right:
+                    queue.append(node.right)
+
+        return depth
 ```
 
 Go：

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
+# Definition for a binary tree node.
-class solution:
+# class TreeNode:
-    def maxdepth(self, root: treenode) -> int:
+#     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()
+        depth = 0
-        queue.append(root)
+        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:
+class Solution:
-    def maxdepth(self, root: 'node') -> int:
+    def maxDepth(self, root: 'Node') -> int:
         if not root:
             return 0
-        depth = 0
+        
-        for i in range(len(root.children)):
+        max_depth = 1
-            depth = max(depth, self.maxdepth(root.children[i]))
+        
-        return 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:
+# Definition for a Node.
-    def maxdepth(self, root: 'node') -> int:
+class Node:
-        queue = collections.deque()
+    def __init__(self, val=None, children=None):
-        if root:
+        self.val = val
-            queue.append(root)
+        self.children = children
-        depth = 0 #记录深度
+"""
+
+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)):
+                for child in node.children:
-                    if node.children[j]:
+                    queue.append(child)
-                        queue.append(node.children[j])
+        
         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:
+# Definition for a Node.
-        st = []
+class Node:
-        if root:
+    def __init__(self, val=None, children=None):
-            st.append(root)
+        self.val = val
-        depth = 0
+        self.children = children
-        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:
+class Solution:
-                node = st.pop()
+    def maxDepth(self, root: 'Node') -> int:
-                depth -= 1
+        if not root:
-            result = max(result, depth)
+            return 0
-        return result
+        
+        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
 ```
 
 

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
+            right_depth = self.minDepth(root.right)
-        return min_depth + 1
+        
+        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()
+        depth = 0
-        que.append(root)
+        queue = collections.deque([root])
-        res = 1
+        
+        while queue:
+            depth += 1 
+            for _ in range(len(queue)):
+                node = queue.popleft()
                 
-        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
                 if not node.left and not node.right:
-                    return res
+                    return depth
-                if node.left is not None:
+            
-                    que.append(node.left)
+                if node.left:
-                if node.right is not None:
+                    queue.append(node.left)
-                    que.append(node.right)
+                    
-            res += 1
+                if node.right:
-        return res
+                    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
 

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
+                # Move one of the pointers back to the start of the list
-                q = slow
+                slow = head
-                while p!=q:
+                while slow != fast:
-                    p = p.next
+                    slow = slow.next
-                    q = q.next
+                    fast = fast.next
-                #你也可以return q
+                return slow
-                return p
+        # 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

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:
+    def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:
-        dummy_head = ListNode(next=head) #添加一个虚拟节点
+        # 创建虚拟头部节点以简化删除过程
-        cur = dummy_head
+        dummy_head = ListNode(next = head)
-        while cur.next:
+        
-            if cur.next.val == val:
+        # 遍历列表并删除值为val的节点
-                cur.next = cur.next.next #删除cur.next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：

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:
-        
+        return self.reverse(head, None)
-        def reverse(pre,cur):
+    def reverse(self, cur: ListNode, pre: ListNode) -> ListNode:
-            if not cur:
+        if cur == None:
-                return pre
+            return pre
-                
+        temp = cur.next
-            tmp = cur.next
+        cur.next = pre
-            cur.next = pre
+        return self.reverse(temp, cur)
-
-            return reverse(cur,tmp)
-        
-        return reverse(None,head)
 
 ```
 
-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：
 

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
+        l = len(nums)
-        Sum = 0     # 滑动窗口数值之和
+        left = 0
-        i = 0      # 滑动窗口起始位置
+        right = 0
-        for j in range(len(nums)):
+        min_len = float('inf')
-            Sum += nums[j]
+        cur_sum = 0 #当前的累加值
-            while Sum >= s:
+        
-                res = min(res, j-i+1)
+        while right < l:
-                Sum -= nums[i]
+            cur_sum += nums[right]
-                i += 1
+            
-        return 0 if res == float("inf") else res
+            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：

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
 

problems/0226.翻转二叉树.md

@@ -314,21 +314,99 @@ 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
+        root.left, root.right = root.right, root.left
-        self.invertTree(root.left) #左
+        self.invertTree(root.left)
-        self.invertTree(root.right) #右
+        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)
@@ -336,59 +414,58 @@ class Solution:
         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
+            return None      
-        st = []
+        stack = [root]        
-        st.append(root)
+        while stack:
-        while st:
+            node = stack.pop()                   
-            node = st.pop()
-            node.left, node.right = node.right, node.left #中
-            if node.right:
-                st.append(node.right) #右
             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 not root: 
-        if root:
+            return None
-            queue.append(root)
+
+        queue = collections.deque([root])    
         while queue:
-            size = len(queue)
+            for i in range(len(queue)):
-            for i in range(size):
                 node = queue.popleft()
-                node.left, node.right = node.right, node.left #节点处理
+                node.left, node.right = node.right, node.left
-                if node.left:
+                if node.left: queue.append(node.left)
-                    queue.append(node.left)
+                if node.right: queue.append(node.right)
-                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)
         return root   
+
 ```
+
 ### Go
 
 递归版本的前序遍历

problems/0707.设计链表.md

@@ -485,172 +485,166 @@ class MyLinkedList {
 
 Python：
 ```python
-# 单链表
+（版本一）单链表法
-class MyLinkedList1:
+class ListNode:
-
+    def __init__(self, val=0, next=None):
-    def __init__(self):
-        self.dummy_head = Node()# 添加虚拟头指针，便于操作
-        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.dummy_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.dummy_head.next
-        self.dummy_head.next = new_node
-        self.size += 1
-
-    def addAtTail(self, val):
-        """
-        :type val: int
-        :rtype: None
-        """
-        new_node = Node(val)
-        cur = self.dummy_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)
-        cur = self.dummy_head
-        while(index):
-            cur = cur.next
-            index -= 1
-        node.next = cur.next
-        cur.next = node
-        self.size += 1
-        
-    def deleteAtIndex(self, index):
-        """
-        :type index: int
-        :rtype: None
-        """
-        if index < 0 or index >= self.size:
-            return
-        pre = self.dummy_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=0, next = None, prev = None):
         self.val = val
         self.next = next
-        self.prev = prev
         
 class MyLinkedList:
-
     def __init__(self):
-        self._head, self._tail = Node(0), Node(0)  # 虚拟节点
+        self.dummy_head = ListNode()
-        self._head.next, self._tail.prev = self._tail, self._head
+        self.size = 0
-        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
-
-
-    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
 
     def get(self, index: int) -> int:
-        """
+        if index < 0 or index >= self.size:
-        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:
             return -1
         
+        current = self.dummy_head.next
+        for i in range(index):
+            current = current.next
+            
+        return current.val
+
     def addAtHead(self, val: int) -> None:
-        """
+        self.dummy_head.next = ListNode(val, self.dummy_head.next)
-        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.size += 1
-        """
-        self._update(self._head, self._head.next, val)
 
     def addAtTail(self, val: int) -> None:
-        """
+        current = self.dummy_head
-        Append a node of value val to the last element of the linked list.
+        while current.next:
-        """
+            current = current.next
-        self._update(self._tail.prev, self._tail, val)
+        current.next = ListNode(val)
+        self.size += 1
 
     def addAtIndex(self, index: int, val: int) -> None:
-        """
+        if index < 0 or index > self.size:
-        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:
             return
-        node = self._get_node(index)
+        
-        self._update(node.prev, node, val)
+        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:
-        """
+        if index < 0 or index >= self.size:
-        Delete the index-th node in the linked list, if the index is valid.
+            return
-        """
+        
-        if 0 <= index < self._count:
+        current = self.dummy_head
-            node = self._get_node(index)
+        for i in range(index):
-            # 计数-1
+            current = current.next
-            self._count -= 1
+        current.next = current.next.next
-            node.prev.next, node.next.prev = node.next, node.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)
+```
+
+
+```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
 
 
 

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)
+        l, r, i = 0, len(nums)-1, len(nums)-1
-        i,j,k = 0,n - 1,n - 1
+        res = [float('inf')] * len(nums) # 需要提前定义列表，存放结果
-        ans = [-1] * n
+        while l <= r:
-        while i <= j:
+            if nums[l] ** 2 < nums[r] ** 2: # 左右边界进行对比，找出最大值
-            lm = nums[i] ** 2
+                res[i] = nums[r] ** 2
-            rm = nums[j] ** 2
+                r -= 1 # 右指针往左移动
-            if lm > rm:
-                ans[k] = lm
-                i += 1
             else:
-                ans[k] = rm
+                res[i] = nums[l] ** 2
-                j -= 1
+                l += 1 # 左指针往右移动
-            k -= 1
+            i -= 1 # 存放结果的指针需要往前平移一位
-        return ans
+        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：

problems/二叉树的迭代遍历.md

@@ -258,6 +258,8 @@ 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:
@@ -279,6 +281,8 @@ 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:

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
+        if not root:
-        result = []
+            return []
 
-        def traversal(root: TreeNode):
+        left = self.preorderTraversal(root.left)
-            if root == None:
+        right = self.preorderTraversal(root.right)
-                return
-            result.append(root.val) # 前序
-            traversal(root.left)    # 左
-            traversal(root.right)   # 右
 
-        traversal(root)
+        return  [root.val] + left +  right
-        return result
 
+```
+```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):
+        left = self.inorderTraversal(root.left)
-            if root == None:
+        right = self.inorderTraversal(root.right)
-                return
+
-            traversal(root.left)    # 左
+        return left + [root.val] + right
-            result.append(root.val) # 中序
+```
-            traversal(root.right)   # 右
+```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):
+        left = self.postorderTraversal(root.left)
-            if root == None:
+        right = self.postorderTraversal(root.right)
-                return
-            traversal(root.left)    # 左
-            traversal(root.right)   # 右
-            result.append(root.val) # 后序
 
-        traversal(root)
+        return left + right + [root.val]
-        return result
 ```
 
 Go：

problems/面试题02.07.链表相交.md

@@ -152,7 +152,7 @@ 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
@@ -178,7 +178,68 @@ 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）等比例法
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+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