Merge branch 'master' of https://github.com/youngyangyang04/leetcode

problems/0017.电话号码的字母组合.md

@@ -137,7 +137,7 @@ for (int i = 0; i < letters.size(); i++) {
 关键地方都讲完了，按照[关于回溯算法，你该了解这些！](https://mp.weixin.qq.com/s/gjSgJbNbd1eAA5WkA-HeWw)中的回溯法模板，不难写出如下C++代码：
 
 
-```
+```c++
 // 版本一
 class Solution {
 private:
@@ -183,7 +183,7 @@ public:
 
 一些写法，是把回溯的过程放在递归函数里了，例如如下代码，我可以写成这样：（注意注释中不一样的地方）
 
-```
+```c++
 // 版本二
 class Solution {
 private:
@@ -319,7 +319,7 @@ class Solution:
 
 python3：
 
-```python3
+```py
 class Solution:
     def letterCombinations(self, digits: str) -> List[str]:
         self.s = ""
@@ -342,6 +342,33 @@ class Solution:
 
 Go：
 
+javaScript：
+
+```js
+var letterCombinations = function(digits) {
+    const k = digits.length;
+    const map = ["","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"];
+    if(!k) return [];
+    if(k === 1) return map[digits].split("");
+
+    const res = [], path = [];
+    backtracking(digits, k, 0);
+    return res;
+
+    function backtracking(n, k, a) {
+        if(path.length === k) {
+            res.push(path.join(""));
+            return;
+        }
+        for(const v of map[n[a]]) {
+            path.push(v);
+            backtracking(n, k, a + 1);
+            path.pop();
+        }
+
+    }
+};
+```
 
 
 

problems/0027.移除元素.md

@@ -199,7 +199,23 @@ def remove_element(nums, val)
     i
 end
 ```
-
+Rust:
+```rust
+pub fn remove_element(nums: &mut Vec<i32>, val: i32) -> &mut Vec<i32> {
+    let mut start: usize = 0;
+    while start < nums.len() {
+        if nums[start] == val {
+            nums.remove(start);
+        }
+        start += 1;
+    }
+    nums
+}
+fn main() {
+    let mut nums = vec![5,1,3,5,2,3,4,1];
+    println!("{:?}",remove_element(&mut nums, 5));
+}
+```
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
 * B站视频：[代码随想录](https://space.bilibili.com/525438321)

problems/0039.组合总和.md

@@ -286,28 +286,28 @@ class Solution:
 ```
 Go：
 
-JavaScript
+JavaScript：
+
 ```js
-var strStr = function (haystack, needle) {
-  if (needle === '') {
-    return 0;
+var combinationSum = function(candidates, target) {
+    const res = [], path = [];
+    candidates.sort(); // 排序
+    backtracking(0, 0);
+    return res;
+    function backtracking(j, sum) {
+        if (sum > target) return;
+        if (sum === target) {
+            res.push(Array.from(path));
+            return;
         }
-
-  let hayslen = haystack.length;
-  let needlen = needle.length;
-
-  if (haystack === '' || hayslen < needlen) {
-    return -1;
-  }
-
-  for (let i = 0; i <= hayslen - needlen; i++) {
-    if (haystack[i] === needle[0]) {
-      if (haystack.substr(i, needlen) === needle) {
-        return i;
-      }
-    }
-    if (i === hayslen - needlen) {
-      return -1;
+        for(let i = j; i < candidates.length; i++ ) {
+            const n = candidates[i];
+            if(n > target - sum) continue;
+            path.push(n);
+            sum += n;
+            backtracking(i, sum);
+            path.pop();
+            sum -= n;
         }
     }
 };

problems/0040.组合总和II.md

@@ -293,7 +293,7 @@ class Solution {
 }
 ```
 Python：
-```python3
+```py
 class Solution:
     def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
         res = []
@@ -314,7 +314,39 @@ class Solution:
 ```
 Go：
 
+javaScript： 
 
+```js
+/**
+ * @param {number[]} candidates
+ * @param {number} target
+ * @return {number[][]}
+ */
+var combinationSum2 = function(candidates, target) {
+    const res = []; path = [], len = candidates.length;
+    candidates.sort();
+    backtracking(0, 0);
+    return res;
+    function backtracking(sum, i) {
+        if (sum > target) return;
+        if (sum === target) {
+            res.push(Array.from(path));
+            return;
+        }
+        let f = -1;
+        for(let j = i; j < len; j++) {
+            const n = candidates[j];
+            if(n > target - sum || n === f) continue;
+            path.push(n);
+            sum += n;
+            f = n;
+            backtracking(sum, j + 1);
+            path.pop();
+            sum -= n;
+        }
+    }
+};
+```
 
 
 -----------------------

problems/0077.组合优化.md

@@ -147,7 +147,7 @@ public:
 
 
 Java：
-```
+```java
 class Solution {
     List<List<Integer>> result = new ArrayList<>();
     LinkedList<Integer> path = new LinkedList<>();
@@ -220,6 +220,28 @@ func backtrack(n,k,start int,track []int){
 }
 ```
 
+javaScript:
+
+```js
+var combine = function(n, k) {
+    const res = [], path = [];
+    backtracking(n, k, 1);
+    return res;
+    function backtracking (n, k, i){
+        const len = path.length;
+        if(len === k) {
+            res.push(Array.from(path));
+            return;
+        }
+        for(let a = i; a <= n + len - k + 1; a++) {
+            path.push(a);
+            backtracking(n, k, a + 1);
+            path.pop();
+        }
+    }
+};
+```
+
 
 
 

problems/0093.复原IP地址.md

@@ -338,6 +338,35 @@ class Solution(object):
         return ans```
 ```
 
+JavaScript： 
+
+```js
+/**
+ * @param {string} s
+ * @return {string[]}
+ */
+var restoreIpAddresses = function(s) {
+    const res = [], path = [];
+    backtracking(0, 0)
+    return res;
+    function backtracking(i) {
+        const len = path.length;
+        if(len > 4) return;
+        if(len === 4 && i === s.length) {
+            res.push(path.join("."));
+            return;
+        }
+        for(let j = i; j < s.length; j++) {
+            const str = s.substr(i, j - i + 1);
+            if(str.length > 3 || +str > 255) break;
+            if(str.length > 1 && str[0] === "0") break;
+            path.push(str);
+            backtracking(j + 1);
+            path.pop()
+        }
+    }
+};
+```
 
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)

problems/0101.对称二叉树.md

@@ -360,6 +360,78 @@ Java：
 
 Python：
 
+> 递归法
+```python
+class Solution:
+    def isSymmetric(self, root: TreeNode) -> bool:
+        if not root:
+            return True
+        return self.compare(root.left, root.right)
+        
+    def compare(self, left, right):
+        #首先排除空节点的情况
+        if left == None and right != None: return False
+        elif left != None and right == None: return False
+        elif left == None and right == None: return True
+        #排除了空节点，再排除数值不相同的情况
+        elif left.val != right.val: return False
+        
+        #此时就是：左右节点都不为空，且数值相同的情况
+        #此时才做递归，做下一层的判断
+        outside = self.compare(left.left, right.right) #左子树：左、 右子树：右
+        inside = self.compare(left.right, right.left) #左子树：右、 右子树：左
+        isSame = outside and inside #左子树：中、 右子树：中 （逻辑处理）
+        return isSame
+```
+
+> 迭代法： 使用队列
+```python
+import collections
+class Solution:
+    def isSymmetric(self, root: TreeNode) -> bool:
+        if not root:
+            return True
+        queue = collections.deque()
+        queue.append(root.left) #将左子树头结点加入队列
+        queue.append(root.right) #将右子树头结点加入队列
+        while queue: #接下来就要判断这这两个树是否相互翻转
+            leftNode = queue.popleft()
+            rightNode = queue.popleft()
+            if not leftNode and not rightNode: #左节点为空、右节点为空，此时说明是对称的
+                continue
+            
+            #左右一个节点不为空，或者都不为空但数值不相同，返回false
+            if not leftNode or not rightNode or leftNode.val != rightNode.val:
+                return False
+            queue.append(leftNode.left) #加入左节点左孩子
+            queue.append(rightNode.right) #加入右节点右孩子
+            queue.append(leftNode.right) #加入左节点右孩子
+            queue.append(rightNode.left) #加入右节点左孩子
+        return True
+```
+
+> 迭代法：使用栈
+```python
+class Solution:
+    def isSymmetric(self, root: TreeNode) -> bool:
+        if not root:
+            return True
+        st = [] #这里改成了栈
+        st.append(root.left)
+        st.append(root.right)
+        while st:
+            leftNode = st.pop()
+            rightNode = st.pop()
+            if not leftNode and not rightNode:
+                continue
+            if not leftNode or not rightNode or leftNode.val != rightNode.val:
+                return False
+            st.append(leftNode.left)
+            st.append(rightNode.right)
+            st.append(leftNode.right)
+            st.append(rightNode.left)
+        return True
+```
 
 Go：
 

problems/0102.二叉树的层序遍历.md

@@ -80,6 +80,40 @@ 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 levelOrder(self, root: TreeNode) -> List[List[int]]:
+        if not root:
+            return []
+
+        quene = [root]
+        out_list = []
+
+        while quene:
+            in_list = []
+            for _ in range(len(quene)):
+                node = quene.pop(0)
+                in_list.append(node.val)
+                if node.left:
+                    quene.append(node.left)
+                if node.right:
+                    quene.append(node.right)
+            
+            out_list.append(in_list)
+
+        return out_list
+```
+
+
+
 javascript代码：
 
 ```javascript
@@ -151,6 +185,43 @@ 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 levelOrderBottom(self, root: TreeNode) -> List[List[int]]:
+        if not root:
+            return []
+        quene = [root]
+        out_list = []
+        
+        while quene:
+            in_list = []
+            for _ in range(len(quene)):
+                node = quene.pop(0)
+                in_list.append(node.val)
+                if node.left:
+                    quene.append(node.left)
+                if node.right:
+                    quene.append(node.right)
+ 
+            out_list.append(in_list)
+    
+        out_list.reverse()
+        return out_list
+    
+# 执行用时：36 ms, 在所有 Python3 提交中击败了92.00%的用户
+# 内存消耗：15.2 MB, 在所有 Python3 提交中击败了63.76%的用户
+```
+
+
+
 javascript代码
 
 ```javascript
@@ -212,6 +283,50 @@ 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()
+                if node.left:
+                    quene.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
+```
+
+
+
+
+
 javascript代码:
 
 ```javascript
@@ -275,6 +390,46 @@ 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 averageOfLevels(self, root: TreeNode) -> List[float]:
+        if not root:
+            return []
+        
+        quene = deque([root])
+        out_list = []
+
+        while quene:
+            in_list = []
+
+            for _ in range(len(quene)):
+                node = quene.popleft()
+                in_list.append(node.val)
+                if node.left:
+                    quene.append(node.left)
+                if node.right:
+                    quene.append(node.right)
+
+            out_list.append(in_list)
+
+        out_list = map(lambda x: sum(x) / len(x), out_list)
+    
+        return out_list
+        
+# 执行用时：56 ms, 在所有 Python3 提交中击败了81.48%的用户
+# 内存消耗：17 MB, 在所有 Python3 提交中击败了89.68%的用户
+```
+
+
+
 javascript代码：
 
 ```javascript
@@ -351,7 +506,68 @@ 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:
+    def levelOrder(self, root: 'Node') -> List[List[int]]:
+        if not root:
+            return []
+        
+        quene = deque([root])
+        out_list = []
+
+        while quene:
+            in_list = []
+
+            for _ in range(len(quene)):
+                node = quene.popleft()
+                in_list.append(node.val)
+                if node.children:
+                    # 这个地方要用extend而不是append，我们看下面的例子：
+                    # In [18]: alist=[]
+                    # In [19]: alist.append([1,2,3])
+                    # In [20]: alist
+                    # Out[20]: [[1, 2, 3]]
+                    # In [21]: alist.extend([4,5,6])
+                    # In [22]: alist
+                    # Out[22]: [[1, 2, 3], 4, 5, 6]
+                    # 可以看到extend对要添加的list进行了一个解包操作
+                    # print(root.children)，可以得到children是一个包含
+                    # 孩子节点地址的list，我们使用for遍历quene的时候，
+                    # 希望quene是一个单层list，所以要用extend
+                    # 使用extend的情况，如果print(quene),结果是
+                    # deque([<__main__.Node object at 0x7f60763ae0a0>])
+				   # deque([<__main__.Node object at 0x7f607636e6d0>, <__main__.Node object at 0x7f607636e130>, <__main__.Node object at 0x7f607636e310>])
+				  # deque([<__main__.Node object at 0x7f607636e880>, <__main__.Node object at 0x7f607636ef10>])
+				  # 可以看到是单层list
+                    # 如果使用append，print(quene)的结果是
+                    # deque([<__main__.Node object at 0x7f18907530a0>])
+				  # deque([[<__main__.Node object at 0x7f18907136d0>, <__main__.Node object at 0x7f1890713130>, <__main__.Node object at 0x7f1890713310>]])
+				  # 可以看到是两层list，这样for的遍历就会报错
+                    
+                    quene.extend(node.children)
+                
+            out_list.append(in_list)
+        
+        return out_list
+    
+# 执行用时：60 ms, 在所有 Python3 提交中击败了76.99%的用户
+# 内存消耗：16.5 MB, 在所有 Python3 提交中击败了89.19%的用户
+```
+
+
+
 JavaScript代码：
+
 ```JavaScript
 var levelOrder = function(root) {
     //每一层可能有2个以上,所以不再使用node.left node.right

problems/0104.二叉树的最大深度.md

@@ -247,6 +247,111 @@ class Solution {
 
 Python：
 
+104.二叉树的最大深度
+> 递归法：
+```python
+class Solution:
+    def maxDepth(self, root: TreeNode) -> int:
+        return self.getDepth(root)
+        
+    def getDepth(self, node):
+        if not node:
+            return 0
+        leftDepth = self.getDepth(node.left) #左
+        rightDepth = self.getDepth(node.right) #右
+        depth = 1 + max(leftDepth, rightDepth) #中
+        return depth
+```
+> 递归法；精简代码
+```python
+class Solution:
+    def maxDepth(self, root: TreeNode) -> int:
+        if not root:
+            return 0
+        return 1 + max(self.maxDepth(root.left), self.maxDepth(root.right))
+```
+
+> 迭代法：
+```python
+import collections
+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)
+        while queue:
+            size = len(queue)
+            depth += 1
+            for i in range(size):
+                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:
+        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
+```
+
+> 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
+        while queue:
+            size = len(queue)
+            depth += 1
+            for i in range(size):
+                node = queue.popleft()
+                for j in range(len(node.children)):
+                    if node.children[j]:
+                        queue.append(node.children[j])
+        return depth
+```
+
+> 使用栈来模拟后序遍历依然可以
+```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
+```
+
 
 Go：
 

problems/0110.平衡二叉树.md

@@ -498,6 +498,62 @@ class Solution {
 
 Python：
 
+> 递归法：
+```python
+class Solution:
+    def isBalanced(self, root: TreeNode) -> bool:
+        return True if self.getDepth(root) != -1 else False
+    
+    #返回以该节点为根节点的二叉树的高度，如果不是二叉搜索树了则返回-1
+    def getDepth(self, node):
+        if not node:
+            return 0
+        leftDepth = self.getDepth(node.left)
+        if leftDepth == -1: return -1 #说明左子树已经不是二叉平衡树
+        rightDepth = self.getDepth(node.right)
+        if rightDepth == -1: return -1 #说明右子树已经不是二叉平衡树
+        return -1 if abs(leftDepth - rightDepth)>1 else 1 + max(leftDepth, rightDepth)
+```
+
+> 迭代法：
+```python
+class Solution:
+    def isBalanced(self, root: TreeNode) -> bool:
+        st = []
+        if not root:
+            return True
+        st.append(root)
+        while st:
+            node = st.pop() #中
+            if abs(self.getDepth(node.left) - self.getDepth(node.right)) > 1:
+                return False
+            if node.right:
+                st.append(node.right) #右（空节点不入栈）
+            if node.left:
+                st.append(node.left) #左（空节点不入栈）
+        return True
+    
+    def getDepth(self, cur):
+        st = []
+        if cur:
+            st.append(cur)
+        depth = 0
+        result = 0
+        while st:
+            node = st.pop()
+            if node:
+                st.append(node) #中
+                st.append(None)
+                depth += 1
+                if node.right: st.append(node.right) #右
+                if node.left: st.append(node.left) #左
+            else:
+                node = st.pop()
+                depth -= 1
+            result = max(result, depth)
+        return result
+```
+
 
 Go：
 ```Go

problems/0131.分割回文串.md

@@ -292,7 +292,7 @@ class Solution {
 ```
 
 Python：
-```python3
+```py
 class Solution:
     def partition(self, s: str) -> List[List[str]]:
         res = []  
@@ -313,7 +313,38 @@ class Solution:
 
 Go：
 
+javaScript：
 
+```js
+/**
+ * @param {string} s
+ * @return {string[][]}
+ */
+const isPalindrome = (s, l, r) => {
+    for (let i = l, j = r; i < j; i++, j--) {
+        if(s[i] !== s[j]) return false;
+    }
+    return true;
+}
+
+var partition = function(s) {
+    const res = [], path = [], len = s.length;
+    backtracking(0);
+    return res;
+    function backtracking(i) {
+        if(i >= len) {
+            res.push(Array.from(path));
+            return;
+        }
+        for(let j = i; j < len; j++) {
+            if(!isPalindrome(s, i, j)) continue;
+            path.push(s.substr(i, j - i + 1));
+            backtracking(j + 1);
+            path.pop();
+        }
+    }
+};
+```
 
 
 -----------------------

problems/0198.打家劫舍.md

@@ -131,7 +131,20 @@ class Solution {
 ```
 
 Python：
-
+```python
+class Solution:
+    def rob(self, nums: List[int]) -> int:
+        if len(nums) == 0:
+            return 0
+        if len(nums) == 1:
+            return nums[0]
+        dp = [0] * len(nums)
+        dp[0] = nums[0]
+        dp[1] = max(nums[0], nums[1])
+        for i in range(2, len(nums)):
+            dp[i] = max(dp[i-2]+nums[i], dp[i-1])
+        return dp[-1]
+```
 
 Go：
 ```Go

problems/0216.组合总和III.md

@@ -180,7 +180,7 @@ if (sum > targetSum) { // 剪枝操作
 
 最后C++代码如下：
 
-```
+```c++
 class Solution {
 private:
     vector<vector<int>> result; // 存放结果集
@@ -262,7 +262,7 @@ class Solution {
 ```
 
 Python：
-```python3
+```py
 class Solution:
     def combinationSum3(self, k: int, n: int) -> List[List[int]]:
         res = []  #存放结果集
@@ -284,6 +284,44 @@ class Solution:
 
 Go：
 
+javaScript:
+
+```js
+// 等差数列
+var maxV = k => k * (9 + 10 - k) / 2;
+var minV = k => k * (1 + k) / 2;
+var combinationSum3 = function(k, n) {
+    if (k > 9 || k < 1) return [];
+    // if (n > maxV(k) || n < minV(k)) return [];
+    // if (n === maxV(k)) return [Array.from({length: k}).map((v, i) => 9 - i)];
+    // if (n === minV(k)) return [Array.from({length: k}).map((v, i) => i + 1)];
+
+    const res = [], path = [];
+    backtracking(k, n, 1, 0);
+    return res;
+    function backtracking(k, n, i, sum){
+        const len = path.length;
+        if (len > k || sum > n) return;
+        if (maxV(k - len) < n - sum) return;
+        if (minV(k - len) > n - sum) return; 
+
+        if(len === k && sum == n) {
+            res.push(Array.from(path));
+            return;
+        }
+
+        const min = Math.min(n - sum, 9 + len - k + 1);
+
+        for(let a = i; a <= min; a++) {
+            path.push(a);
+            sum += a;
+            backtracking(k, n, a + 1, sum);
+            path.pop();
+            sum -= a;
+        }
+    }
+};
+```
 
 
 

problems/0222.完全二叉树的节点个数.md

@@ -240,9 +240,103 @@ class Solution {
 
 Python：
 
+> 递归法：
+```python
+class Solution:
+    def countNodes(self, root: TreeNode) -> int:
+        return self.getNodesNum(root)
+        
+    def getNodesNum(self, cur):
+        if not cur:
+            return 0
+        leftNum = self.getNodesNum(cur.left) #左
+        rightNum = self.getNodesNum(cur.right) #右
+        treeNum = leftNum + rightNum + 1 #中
+        return treeNum
+```
+
+> 递归法：精简版
+```python
+class Solution:
+    def countNodes(self, root: TreeNode) -> int:
+        if not root:
+            return 0
+        return 1 + self.countNodes(root.left) + self.countNodes(root.right)
+```
+
+> 迭代法：
+```python
+import collections
+class Solution:
+    def countNodes(self, root: TreeNode) -> int:
+        queue = collections.deque()
+        if root:
+            queue.append(root)
+        result = 0
+        while queue:
+            size = len(queue)
+            for i in range(size):
+                node = queue.popleft()
+                result += 1 #记录节点数量
+                if node.left:
+                    queue.append(node.left)
+                if node.right:
+                    queue.append(node.right)
+        return result
+```
+
+> 完全二叉树
+```python
+class Solution:
+    def countNodes(self, root: TreeNode) -> int:
+        if not root:
+            return 0
+        left = root.left
+        right = root.right
+        leftHeight = 0 #这里初始为0是有目的的，为了下面求指数方便
+        rightHeight = 0
+        while left: #求左子树深度
+            left = left.left
+            leftHeight += 1
+        while right: #求右子树深度
+            right = right.right
+            rightHeight += 1
+        if leftHeight == rightHeight:
+            return (2 << leftHeight) - 1 #注意(2<<1) 相当于2^2，所以leftHeight初始为0
+        return self.countNodes(root.left) + self.countNodes(root.right) + 1
+```
 
 Go：
 
+递归版本
+
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+//本题直接就是求有多少个节点，无脑存进数组算长度就行了。
+func countNodes(root *TreeNode) int {
+    if root == nil {
+        return 0
+    }
+    res := 1
+    if root.Right != nil {
+        res += countNodes(root.Right)
+    }
+    if root.Left != nil {
+        res += countNodes(root.Left)
+    }
+    return res
+}
+```
+
+
+
 JavaScript:
 
 递归版本

problems/0226.翻转二叉树.md

@@ -230,6 +230,56 @@ class Solution {
 
 Python：
 
+> 递归法：前序遍历
+```python
+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
+        return root
+```
+
+> 迭代法：深度优先遍历（前序遍历）
+```python
+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
+            if node.left:
+                st.append(node.left) #左
+        return root
+```
+
+> 迭代法：广度优先遍历（层序遍历）
+```python
+import collections
+class Solution:
+    def invertTree(self, root: TreeNode) -> TreeNode:
+        queue = collections.deque() #使用deque()
+        if root:
+            queue.append(root)
+        while queue:
+            size = len(queue)
+            for i in range(size):
+                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
+```
+
 Go：
 ```Go
 func invertTree(root *TreeNode) *TreeNode {

problems/0337.打家劫舍III.md

@@ -287,6 +287,25 @@ class Solution {
 
 Python：
 
+> 动态规划
+```python
+class Solution:
+    def rob(self, root: TreeNode) -> int:
+        result = self.robTree(root)
+        return max(result[0], result[1])
+    
+    #长度为2的数组，0：不偷，1：偷
+    def robTree(self, cur):
+        if not cur:
+            return (0, 0) #这里返回tuple, 也可以返回list
+        left = self.robTree(cur.left)
+        right = self.robTree(cur.right)
+        #偷cur
+        val1 = cur.val + left[0] + right[0]
+        #不偷cur
+        val2 = max(left[0], left[1]) + max(right[0], right[1])
+        return (val2, val1)
+```
 
 Go：
 

problems/0377.组合总和Ⅳ.md

@@ -163,7 +163,7 @@ class Solution {
         return dp[target];
     }
 }
-
+```
 
 Python：
 

problems/0474.一和零.md

@@ -190,10 +190,59 @@ class Solution {
 ```
 
 Python：
-
+```python3
+class Solution:
+    def findMaxForm(self, strs: List[str], m: int, n: int) -> int:
+        dp = [[0] * (n + 1) for _ in range(m + 1)]	# 默认初始化0
+        # 遍历物品
+        for str in strs:
+            ones = str.count('1')
+            zeros = str.count('0')
+            # 遍历背包容量且从后向前遍历！
+            for i in range(m, zeros - 1, -1):
+                for j in range(n, ones - 1, -1):
+                    dp[i][j] = max(dp[i][j], dp[i - zeros][j - ones] + 1)
+        return dp[m][n]
+```
 
 Go：
+```go
+func findMaxForm(strs []string, m int, n int) int {
+	// 定义数组
+	dp := make([][]int, m+1)
+	for i,_ := range dp {
+		dp[i] = make([]int, n+1 )
+	}
+	// 遍历
+	for i:=0;i<len(strs);i++ {
+		zeroNum,oneNum := 0 , 0
+		//计算0,1 个数
+		//或者直接strings.Count(strs[i],"0")
+		for _,v := range strs[i] {
+			if v == '0' {
+				zeroNum++
+			}
+		}
+		oneNum = len(strs[i])-zeroNum
+		// 从后往前 遍历背包容量
+		for j:= m ; j >= zeroNum;j-- {
+			for k:=n ; k >= oneNum;k-- {
+				// 推导公式
+				dp[j][k] = max(dp[j][k],dp[j-zeroNum][k-oneNum]+1)
+			}
+		}
+		//fmt.Println(dp)
+	}
+	return dp[m][n]
+}
 
+func max(a,b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+```
 
 
 

problems/0530.二叉搜索树的最小绝对差.md

@@ -151,7 +151,30 @@ public:
 
 
 Java：
-
+递归
+```java
+class Solution {
+    TreeNode pre;// 记录上一个遍历的结点
+    int result = Integer.MAX_VALUE;
+    public int getMinimumDifference(TreeNode root) {
+       if(root==null)return 0;
+       traversal(root);
+       return result;
+    }
+    public void traversal(TreeNode root){
+        if(root==null)return;
+        //左
+        traversal(root.left);
+        //中
+        if(pre!=null){
+            result = Math.min(result,root.val-pre.val);
+        }
+        pre = root;
+        //右
+        traversal(root.right);
+    }
+}
+```
 ```Java
 class Solution {
     TreeNode pre;// 记录上一个遍历的结点

problems/0541.反转字符串II.md

@@ -164,6 +164,30 @@ class Solution(object):
 
 
 Go：
+```go
+func reverseStr(s string, k int) string {
+    ss := []byte(s)
+    length := len(s)
+    for i := 0; i < length; i += 2 * k {
+        if i + k <= length {
+            reverse(ss[i:i+k])
+        } else {
+            reverse(ss[i:length])
+        }
+    }
+    return string(ss)
+}
+
+func reverse(b []byte) {
+    left := 0
+    right := len(b) - 1
+    for left < right {
+        b[left], b[right] = b[right], b[left]
+        left++
+        right--
+    }
+}
+```
 
 javaScript: 
 

problems/0674.最长连续递增序列.md

@@ -156,7 +156,31 @@ public:
 
 
 Java：
-
+```java
+ /**
+     * 1.dp[i] 代表当前下表最大连续值
+     * 2.递推公式 if（nums[i+1]>nums[i]） dp[i+1] = dp[i]+1
+     * 3.初始化 都为1
+     * 4.遍历方向，从其那往后
+     * 5.结果推导 。。。。
+     * @param nums
+     * @return
+     */
+    public static int findLengthOfLCIS(int[] nums) {
+        int[] dp = new int[nums.length];
+        for (int i = 0; i < dp.length; i++) {
+            dp[i] = 1;
+        }
+        int res = 1;
+        for (int i = 0; i < nums.length - 1; i++) {
+            if (nums[i + 1] > nums[i]) {
+                dp[i + 1] = dp[i] + 1;
+            }
+            res = res > dp[i + 1] ? res : dp[i + 1];
+        }
+        return res;
+    }
+```
 
 Python：
 

problems/1005.K次取反后最大化的数组和.md

@@ -102,7 +102,7 @@ Java：
 ```java
 class Solution {
     public int largestSumAfterKNegations(int[] A, int K) {
-        if (A.length == 1) return A[0];
+        if (A.length == 1) return k % 2 == 0 ? A[0] : -A[0];
         Arrays.sort(A);
         int sum = 0;
         int idx = 0;

problems/二叉树中递归带着回溯.md

@@ -257,6 +257,83 @@ Java：
 
 Python：
 
+100.相同的树
+> 递归法
+```python
+class Solution:
+    def isSameTree(self, p: TreeNode, q: TreeNode) -> bool:
+        return self.compare(p, q)
+        
+    def compare(self, tree1, tree2):
+        if not tree1 and tree2:
+            return False
+        elif tree1 and not tree2:
+            return False
+        elif not tree1 and not tree2:
+            return True
+        elif tree1.val != tree2.val: #注意这里我没有使用else
+            return False
+        
+        #此时就是：左右节点都不为空，且数值相同的情况
+        #此时才做递归，做下一层的判断
+        compareLeft = self.compare(tree1.left, tree2.left) #左子树：左、 右子树：左
+        compareRight = self.compare(tree1.right, tree2.right) #左子树：右、 右子树：右
+        isSame = compareLeft and compareRight #左子树：中、 右子树：中（逻辑处理）
+        return isSame
+```
+
+257.二叉的所有路径
+> 递归中隐藏着回溯
+```python
+class Solution:
+    def binaryTreePaths(self, root: TreeNode) -> List[str]:
+        result = []
+        path = []
+        if not root:
+            return result
+        self.traversal(root, path, result)
+        return result
+        
+    def traversal(self, cur, path, result):
+        path.append(cur.val)
+        #这才到了叶子节点
+        if not cur.left and not cur.right:
+            sPath = ""
+            for i in range(len(path)-1):
+                sPath += str(path[i])
+                sPath += "->"
+            sPath += str(path[len(path)-1])
+            result.append(sPath)
+            return
+        if cur.left:
+            self.traversal(cur.left, path, result)
+            path.pop() #回溯
+        if cur.right:
+            self.traversal(cur.right, path, result)
+            path.pop() #回溯
+```
+
+> 精简版
+```python
+class Solution:
+    def binaryTreePaths(self, root: TreeNode) -> List[str]:
+        result = []
+        path = ""
+        if not root:
+            return result
+        self.traversal(root, path, result)
+        return result
+        
+    def traversal(self, cur, path, result):
+        path += str(cur.val) #中
+        if not cur.left and not cur.right:
+            result.append(path)
+            return
+        if cur.left:
+            self.traversal(cur.left, path+"->", result) #左  回溯就隐藏在这里
+        if cur.right:
+            self.traversal(cur.right, path+"->", result) #右 回溯就隐藏在这里
+```
 
 Go：
 

problems/背包理论基础01背包-1.md

@@ -273,7 +273,40 @@ Python：
 
 
 Go：
+```go
+func test_2_wei_bag_problem1(weight, value []int, bagWeight int) int {
+	// 定义dp数组
+	dp := make([][]int, len(weight))
+	for i, _ := range dp {
+		dp[i] = make([]int, bagWeight+1)
+	}
+	// 初始化
+	for j := bagWeight; j >= weight[0]; j-- {
+		dp[0][j] = dp[0][j-weight[0]] + value[0]
+	}
+	// 递推公式
+	for i := 1; i < len(weight); i++ {
+		//正序,也可以倒序
+		for  j := weight[i];j<= bagWeight ; j++ {
+			dp[i][j] = max(dp[i-1][j], dp[i-1][j-weight[i]]+value[i])
+		}		
+	}
+	return dp[len(weight)-1][bagWeight]
+}
 
+func max(a,b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+func main() {
+	weight := []int{1,3,4}
+	value := []int{15,20,30}
+	test_2_wei_bag_problem1(weight,value,4)
+}
+```
 
 -----------------------
 * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)

problems/背包理论基础01背包-2.md

@@ -219,7 +219,36 @@ Python：
 
 
 Go：
+```go
+func test_1_wei_bag_problem(weight, value []int, bagWeight int) int {
+	// 定义 and 初始化
+	dp := make([]int,bagWeight+1)
+	// 递推顺序
+	for i := 0 ;i < len(weight) ; i++ {
+		// 这里必须倒序,区别二维,因为二维dp保存了i的状态
+		for j:= bagWeight; j >= weight[i] ; j-- {
+			// 递推公式
+			dp[j] = max(dp[j], dp[j-weight[i]]+value[i])
+		}
+	}
+	//fmt.Println(dp)
+	return dp[bagWeight]
+}
 
+func max(a,b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
+
+
+func main() {
+	weight := []int{1,3,4}
+	value := []int{15,20,30}
+	test_1_wei_bag_problem(weight,value,4)
+}
+```
 
 
 

problems/背包问题理论基础完全背包.md

@@ -179,9 +179,45 @@ int main() {
 
 Java：
 
-
 Python：
 
+```python3
+# 先遍历物品，再遍历背包
+def test_complete_pack1():
+    weight = [1, 3, 4]
+    value = [15, 20, 30]
+    bag_weight = 4
+
+    dp = [0]*(bag_weight + 1)
+
+    for i in range(len(weight)):
+        for j in range(weight[i], bag_weight + 1):
+            dp[j] = max(dp[j], dp[j - weight[i]] + value[i])
+    
+    print(dp[bag_weight])
+
+# 先遍历背包，再遍历物品
+def test_complete_pack2():
+    weight = [1, 3, 4]
+    value = [15, 20, 30]
+    bag_weight = 4
+
+    dp = [0]*(bag_weight + 1)
+
+    for j in range(bag_weight + 1):
+        for i in range(len(weight)):
+            if j >= weight[i]: dp[j] = max(dp[j], dp[j - weight[i]] + value[i])
+    
+    print(dp[bag_weight])
+
+
+if __name__ == '__main__':
+    test_complete_pack1()
+    test_complete_pack2()
+```
+
+
+
 
 Go：