Merge branch 'master' into master

README.md

@@ -5,10 +5,11 @@
 
 > 1. **介绍**：本项目是一套完整的刷题计划，旨在帮助大家少走弯路，循序渐进学算法，[关注作者](#关于作者)
 > 2. **PDF版本** ： [「代码随想录」算法精讲 PDF 版本](https://programmercarl.com/other/algo_pdf.html) 。
-> 3. **刷题顺序** ： README已经将刷题顺序排好了，按照顺序一道一道刷就可以。
+> 3. **最强八股文：**：[代码随想录知识星球精华PDF](https://www.programmercarl.com/other/kstar_baguwen.html)
-> 4. **学习社区** : 一起学习打卡/面试技巧/如何选择offer/大厂内推/职场规则/简历修改/技术分享/程序人生。欢迎加入[「代码随想录」知识星球](https://programmercarl.com/other/kstar.html) 。
+> 4. **刷题顺序** ： README已经将刷题顺序排好了，按照顺序一道一道刷就可以。
-> 5. **提交代码**：本项目统一使用C++语言进行讲解，但已经有Java、Python、Go、JavaScript等等多语言版本，感谢[这里的每一位贡献者](https://github.com/youngyangyang04/leetcode-master/graphs/contributors)，如果你也想贡献代码点亮你的头像，[点击这里](https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A)了解提交代码的方式。
+> 5. **学习社区** : 一起学习打卡/面试技巧/如何选择offer/大厂内推/职场规则/简历修改/技术分享/程序人生。欢迎加入[「代码随想录」知识星球](https://programmercarl.com/other/kstar.html) 。
-> 6. **转载须知** ：以下所有文章皆为我（[程序员Carl](https://github.com/youngyangyang04)）的原创。引用本项目文章请注明出处，发现恶意抄袭或搬运，会动用法律武器维护自己的权益。让我们一起维护一个良好的技术创作环境！
+> 6. **提交代码**：本项目统一使用C++语言进行讲解，但已经有Java、Python、Go、JavaScript等等多语言版本，感谢[这里的每一位贡献者](https://github.com/youngyangyang04/leetcode-master/graphs/contributors)，如果你也想贡献代码点亮你的头像，[点击这里](https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A)了解提交代码的方式。
+> 7. **转载须知** ：以下所有文章皆为我（[程序员Carl](https://github.com/youngyangyang04)）的原创。引用本项目文章请注明出处，发现恶意抄袭或搬运，会动用法律武器维护自己的权益。让我们一起维护一个良好的技术创作环境！
 
 <p align="center">
 <a href="programmercarl.com" target="_blank">

problems/0045.跳跃游戏II.md

@@ -142,6 +142,7 @@ public:
 
 ### Java 
 ```Java
+// 版本一
 class Solution {
     public int jump(int[] nums) {
         if (nums == null || nums.length == 0 || nums.length == 1) {
@@ -172,7 +173,30 @@ class Solution {
 }
 ```
 
+```java
+// 版本二
+class Solution {
+    public int jump(int[] nums) {
+        int result = 0;
+        // 当前覆盖的最远距离下标
+        int end = 0;
+        // 下一步覆盖的最远距离下标
+        int temp = 0;
+        for (int i = 0; i <= end && end < nums.length - 1; ++i) {
+            temp = Math.max(temp, i + nums[i]);
+            // 可达位置的改变次数就是跳跃次数
+            if (i == end) {
+                end = temp;
+                result++;
+            }
+        }
+        return result;
+    }
+}
+```
+
 ### Python
+
 ```python
 class Solution:
     def jump(self, nums: List[int]) -> int:

problems/0063.不同路径II.md

@@ -4,7 +4,7 @@
 </a>
 <p align="center"><strong><a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
 
-## 63. 不同路径 II
+# 63. 不同路径 II
 
 [力扣题目链接](https://leetcode-cn.com/problems/unique-paths-ii/)
 
@@ -22,23 +22,22 @@
 
 ![](https://img-blog.csdnimg.cn/20210111204939971.png)
 
-输入：obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
+* 输入：obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
-输出：2
+* 输出：2
 解释：
-3x3 网格的正中间有一个障碍物。
+* 3x3 网格的正中间有一个障碍物。
-从左上角到右下角一共有 2 条不同的路径：
+* 从左上角到右下角一共有 2 条不同的路径：
-1. 向右 -> 向右 -> 向下 -> 向下
+    1. 向右 -> 向右 -> 向下 -> 向下
-2. 向下 -> 向下 -> 向右 -> 向右
+    2. 向下 -> 向下 -> 向右 -> 向右
 
 示例 2：
 
 ![](https://img-blog.csdnimg.cn/20210111205857918.png)
 
-输入：obstacleGrid = [[0,1],[0,0]]
+* 输入：obstacleGrid = [[0,1],[0,0]]
-输出：1
+* 输出：1
 
 提示：
-
 * m == obstacleGrid.length
 * n == obstacleGrid[i].length
 * 1 <= m, n <= 100
@@ -171,7 +170,7 @@ public:
 
 ## 其他语言版本
 
-Java：
+### Java 
 
 ```java
 class Solution {
@@ -199,7 +198,7 @@ class Solution {
 ```
 
 
-Python：
+### Python
 
 ```python
 class Solution:
@@ -262,7 +261,7 @@ class Solution:
 ```
 
 
-Go：
+### Go
 
 ```go
 func uniquePathsWithObstacles(obstacleGrid [][]int) int {
@@ -295,8 +294,8 @@ func uniquePathsWithObstacles(obstacleGrid [][]int) int {
 
 ```
 
-Javascript
+### Javascript
-``` Javascript
+```Javascript
 var uniquePathsWithObstacles = function(obstacleGrid) {
     const m = obstacleGrid.length
     const n = obstacleGrid[0].length

problems/0096.不同的二叉搜索树.md

@@ -4,7 +4,7 @@
 </a>
 <p align="center"><strong><a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
 
-## 96.不同的二叉搜索树
+# 96.不同的二叉搜索树
 
 [力扣题目链接](https://leetcode-cn.com/problems/unique-binary-search-trees/)
 
@@ -163,7 +163,7 @@ public:
 ## 其他语言版本
 
 
-Java：
+### Java 
 ```Java
 class Solution {
     public int numTrees(int n) {
@@ -184,7 +184,7 @@ class Solution {
 }
 ```
 
-Python：
+### Python
 ```python
 class Solution:
     def numTrees(self, n: int) -> int:
@@ -196,7 +196,7 @@ class Solution:
         return dp[-1]
 ```
 
-Go：
+### Go
 ```Go
 func numTrees(n int)int{
 	dp:=make([]int,n+1)
@@ -210,7 +210,7 @@ func numTrees(n int)int{
 }
 ```
 
-Javascript：
+### Javascript
 ```Javascript
 const numTrees =(n) => {
     let dp = new Array(n+1).fill(0);

problems/0101.对称二叉树.md

@@ -574,6 +574,75 @@ var isSymmetric = function(root) {
 };
 ```
 
+## TypeScript：
+
+> 递归法
+
+```typescript
+function isSymmetric(root: TreeNode | null): boolean {
+    function recur(node1: TreeNode | null, node2: TreeNode | null): boolean {
+        if (node1 === null && node2 === null) return true;
+        if (node1 === null || node2 === null) return false;
+        if (node1.val !== node2.val) return false
+        let isSym1: boolean = recur(node1.left, node2.right);
+        let isSym2: boolean = recur(node1.right, node2.left);
+        return isSym1 && isSym2
+    }
+    if (root === null) return true;
+    return recur(root.left, root.right);
+};
+```
+
+> 迭代法
+
+```typescript
+// 迭代法(队列)
+function isSymmetric(root: TreeNode | null): boolean {
+    let helperQueue: (TreeNode | null)[] = [];
+    let tempNode1: TreeNode | null,
+        tempNode2: TreeNode | null;
+    if (root !== null) {
+        helperQueue.push(root.left);
+        helperQueue.push(root.right);
+    }
+    while (helperQueue.length > 0) {
+        tempNode1 = helperQueue.shift()!;
+        tempNode2 = helperQueue.shift()!;
+        if (tempNode1 === null && tempNode2 === null) continue;
+        if (tempNode1 === null || tempNode2 === null) return false;
+        if (tempNode1.val !== tempNode2.val) return false;
+        helperQueue.push(tempNode1.left);
+        helperQueue.push(tempNode2.right);
+        helperQueue.push(tempNode1.right);
+        helperQueue.push(tempNode2.left);
+    }
+    return true;
+}
+
+// 迭代法(栈)
+function isSymmetric(root: TreeNode | null): boolean {
+    let helperStack: (TreeNode | null)[] = [];
+    let tempNode1: TreeNode | null,
+        tempNode2: TreeNode | null;
+    if (root !== null) {
+        helperStack.push(root.left);
+        helperStack.push(root.right);
+    }
+    while (helperStack.length > 0) {
+        tempNode1 = helperStack.pop()!;
+        tempNode2 = helperStack.pop()!;
+        if (tempNode1 === null && tempNode2 === null) continue;
+        if (tempNode1 === null || tempNode2 === null) return false;
+        if (tempNode1.val !== tempNode2.val) return false;
+        helperStack.push(tempNode1.left);
+        helperStack.push(tempNode2.right);
+        helperStack.push(tempNode1.right);
+        helperStack.push(tempNode2.left);
+    }
+    return true;
+};
+```
+
 ## Swift:
 
 > 递归

problems/0102.二叉树的层序遍历.md

@@ -246,7 +246,35 @@ var levelOrder = function(root) {
 
 ```
 
+TypeScript：
+
+```typescript
+function levelOrder(root: TreeNode | null): number[][] {
+    let helperQueue: TreeNode[] = [];
+    let res: number[][] = [];
+    let tempArr: number[] = [];
+    if (root !== null) helperQueue.push(root);
+    let curNode: TreeNode;
+    while (helperQueue.length > 0) {
+        for (let i = 0, length = helperQueue.length; i < length; i++) {
+            curNode = helperQueue.shift()!;
+            tempArr.push(curNode.val);
+            if (curNode.left !== null) {
+                helperQueue.push(curNode.left);
+            }
+            if (curNode.right !== null) {
+                helperQueue.push(curNode.right);
+            }
+        }
+        res.push(tempArr);
+        tempArr = [];
+    }
+    return res;
+};
+```
+
 Swift:
+
 ```swift
 func levelOrder(_ root: TreeNode?) -> [[Int]] {
     var res = [[Int]]()
@@ -454,7 +482,31 @@ var levelOrderBottom = function(root) {
 };
 ```
 
+TypeScript:
+
+```typescript
+function levelOrderBottom(root: TreeNode | null): number[][] {
+    let helperQueue: TreeNode[] = [];
+    let resArr: number[][] = [];
+    let tempArr: number[] = [];
+    let tempNode: TreeNode;
+    if (root !== null) helperQueue.push(root);
+    while (helperQueue.length > 0) {
+        for (let i = 0, length = helperQueue.length; i < length; i++) {
+            tempNode = helperQueue.shift()!;
+            tempArr.push(tempNode.val);
+            if (tempNode.left !== null) helperQueue.push(tempNode.left);
+            if (tempNode.right !== null) helperQueue.push(tempNode.right);
+        }
+        resArr.push(tempArr);
+        tempArr = [];
+    }
+    return resArr.reverse();
+};
+```
+
 Swift:
+
 ```swift
 func levelOrderBottom(_ root: TreeNode?) -> [[Int]] {
     var res =  [[Int]]()
@@ -657,7 +709,28 @@ var rightSideView = function(root) {
 };
 ```
 
+TypeScript：
+
+```typescript
+function rightSideView(root: TreeNode | null): number[] {
+    let helperQueue: TreeNode[] = [];
+    let resArr: number[] = [];
+    let tempNode: TreeNode;
+    if (root !== null) helperQueue.push(root);
+    while (helperQueue.length > 0) {
+        for (let i = 0, length = helperQueue.length; i < length; i++) {
+            tempNode = helperQueue.shift()!;
+            if (i === length - 1) resArr.push(tempNode.val);
+            if (tempNode.left !== null) helperQueue.push(tempNode.left);
+            if (tempNode.right !== null) helperQueue.push(tempNode.right);
+        }
+    }
+    return resArr;
+};
+```
+
 Swift:
+
 ```swift
 func rightSideView(_ root: TreeNode?) -> [Int] {
     var res = [Int]()
@@ -868,7 +941,32 @@ var averageOfLevels = function(root) {
 };
 ```
 
+TypeScript：
+
+```typescript
+function averageOfLevels(root: TreeNode | null): number[] {
+    let helperQueue: TreeNode[] = [];
+    let resArr: number[] = [];
+    let total: number = 0;
+    let tempNode: TreeNode;
+    if (root !== null) helperQueue.push(root);
+    while (helperQueue.length > 0) {
+        let length = helperQueue.length;
+        for (let i = 0; i < length; i++) {
+            tempNode = helperQueue.shift()!;
+            total += tempNode.val;
+            if (tempNode.left) helperQueue.push(tempNode.left);
+            if (tempNode.right) helperQueue.push(tempNode.right);
+        }
+        resArr.push(total / length);
+        total = 0;
+    }
+    return resArr;
+};
+```
+
 Swift:
+
 ```swift
 func averageOfLevels(_ root: TreeNode?) -> [Double] {
     var res = [Double]()
@@ -1092,7 +1190,30 @@ var levelOrder = function(root) {
 };
 ```
 
+TypeScript:
+
+```typescript
+function levelOrder(root: Node | null): number[][] {
+    let helperQueue: Node[] = [];
+    let resArr: number[][] = [];
+    let tempArr: number[] = [];
+    if (root !== null) helperQueue.push(root);
+    let curNode: Node;
+    while (helperQueue.length > 0) {
+        for (let i = 0, length = helperQueue.length; i < length; i++) {
+            curNode = helperQueue.shift()!;
+            tempArr.push(curNode.val);
+            helperQueue.push(...curNode.children);
+        }
+        resArr.push(tempArr);
+        tempArr = [];
+    }
+    return resArr;
+};
+```
+
 Swift:
+
 ```swift
 func levelOrder(_ root: Node?) -> [[Int]] {
     var res = [[Int]]()
@@ -1272,7 +1393,34 @@ var largestValues = function(root) {
 };
 ```
 
+TypeScript:
+
+```typescript
+function largestValues(root: TreeNode | null): number[] {
+    let helperQueue: TreeNode[] = [];
+    let resArr: number[] = [];
+    let tempNode: TreeNode;
+    let max: number = 0;
+    if (root !== null) helperQueue.push(root);
+    while (helperQueue.length > 0) {
+        for (let i = 0, length = helperQueue.length; i < length; i++) {
+            tempNode = helperQueue.shift()!;
+            if (i === 0) {
+                max = tempNode.val;
+            } else {
+                max = max > tempNode.val ? max : tempNode.val;
+            }
+            if (tempNode.left) helperQueue.push(tempNode.left);
+            if (tempNode.right) helperQueue.push(tempNode.right);
+        }
+        resArr.push(max);
+    }
+    return resArr;
+};
+```
+
 Swift:
+
 ```swift
 func largestValues(_ root: TreeNode?) -> [Int] {
     var res = [Int]()
@@ -1463,6 +1611,31 @@ var connect = function(root) {
 };
 
 ```
+TypeScript:
+
+```typescript
+function connect(root: Node | null): Node | null {
+    let helperQueue: Node[] = [];
+    let preNode: Node, curNode: Node;
+    if (root !== null) helperQueue.push(root);
+    while (helperQueue.length > 0) {
+        for (let i = 0, length = helperQueue.length; i < length; i++) {
+            if (i === 0) {
+                preNode = helperQueue.shift()!;
+            } else {
+                curNode = helperQueue.shift()!;
+                preNode.next = curNode;
+                preNode = curNode;
+            }
+            if (preNode.left) helperQueue.push(preNode.left);
+            if (preNode.right) helperQueue.push(preNode.right);
+        }
+        preNode.next = null;
+    }
+    return root;
+};
+```
+
 go:
 
 ```GO
@@ -1689,6 +1862,31 @@ var connect = function(root) {
     return root;
 };
 ```
+TypeScript:
+
+```typescript
+function connect(root: Node | null): Node | null {
+    let helperQueue: Node[] = [];
+    let preNode: Node, curNode: Node;
+    if (root !== null) helperQueue.push(root);
+    while (helperQueue.length > 0) {
+        for (let i = 0, length = helperQueue.length; i < length; i++) {
+            if (i === 0) {
+                preNode = helperQueue.shift()!;
+            } else {
+                curNode = helperQueue.shift()!;
+                preNode.next = curNode;
+                preNode = curNode;
+            }
+            if (preNode.left) helperQueue.push(preNode.left);
+            if (preNode.right) helperQueue.push(preNode.right);
+        }
+        preNode.next = null;
+    }
+    return root;
+};
+```
+
 go:
 
 ```GO
@@ -1933,7 +2131,28 @@ var maxDepth = function(root) {
 };
 ```
 
+TypeScript:
+
+```typescript
+function maxDepth(root: TreeNode | null): number {
+    let helperQueue: TreeNode[] = [];
+    let resDepth: number = 0;
+    let tempNode: TreeNode;
+    if (root !== null) helperQueue.push(root);
+    while (helperQueue.length > 0) {
+        resDepth++;
+        for (let i = 0, length = helperQueue.length; i < length; i++) {
+            tempNode = helperQueue.shift()!;
+            if (tempNode.left) helperQueue.push(tempNode.left);
+            if (tempNode.right) helperQueue.push(tempNode.right);
+        }
+    }
+    return resDepth;
+};
+```
+
 Swift:
+
 ```swift
 func maxDepth(_ root: TreeNode?) -> Int {
     guard let root = root else {
@@ -2130,7 +2349,29 @@ var minDepth = function(root) {
 };
 ```
 
+TypeScript:
+
+```typescript
+function minDepth(root: TreeNode | null): number {
+    let helperQueue: TreeNode[] = [];
+    let resMin: number = 0;
+    let tempNode: TreeNode;
+    if (root !== null) helperQueue.push(root);
+    while (helperQueue.length > 0) {
+        resMin++;
+        for (let i = 0, length = helperQueue.length; i < length; i++) {
+            tempNode = helperQueue.shift()!;
+            if (tempNode.left === null && tempNode.right === null) return resMin;
+            if (tempNode.left !== null) helperQueue.push(tempNode.left);
+            if (tempNode.right !== null) helperQueue.push(tempNode.right);
+        }
+    }
+    return resMin;
+};
+```
+
 Swift：
+
 ```swift
 func minDepth(_ root: TreeNode?) -> Int {
     guard let root = root else {

problems/0104.二叉树的最大深度.md

@@ -598,7 +598,81 @@ var maxDepth = function(root) {
 };
 ```
 
+## TypeScript：
+
+> 二叉树的最大深度：
+
+```typescript
+// 后续遍历（自下而上）
+function maxDepth(root: TreeNode | null): number {
+    if (root === null) return 0;
+    return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
+};
+
+// 前序遍历(自上而下)
+function maxDepth(root: TreeNode | null): number {
+    function recur(node: TreeNode | null, count: number) {
+        if (node === null) {
+            resMax = resMax > count ? resMax : count;
+            return;
+        }
+        recur(node.left, count + 1);
+        recur(node.right, count + 1);
+    }
+    let resMax: number = 0;
+    let count: number = 0;
+    recur(root, count);
+    return resMax;
+};
+
+// 层序遍历（迭代法）
+function maxDepth(root: TreeNode | null): number {
+    let helperQueue: TreeNode[] = [];
+    let resDepth: number = 0;
+    let tempNode: TreeNode;
+    if (root !== null) helperQueue.push(root);
+    while (helperQueue.length > 0) {
+        resDepth++;
+        for (let i = 0, length = helperQueue.length; i < length; i++) {
+            tempNode = helperQueue.shift()!;
+            if (tempNode.left) helperQueue.push(tempNode.left);
+            if (tempNode.right) helperQueue.push(tempNode.right);
+        }
+    }
+    return resDepth;
+};
+```
+
+> N叉树的最大深度
+
+```typescript
+// 后续遍历（自下而上）
+function maxDepth(root: TreeNode | null): number {
+    if (root === null) return 0;
+    return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
+};
+
+// 前序遍历(自上而下)
+function maxDepth(root: TreeNode | null): number {
+    function recur(node: TreeNode | null, count: number) {
+        if (node === null) {
+            resMax = resMax > count ? resMax : count;
+            return;
+        }
+        recur(node.left, count + 1);
+        recur(node.right, count + 1);
+    }
+    let resMax: number = 0;
+    let count: number = 0;
+    recur(root, count);
+    return resMax;
+};
+
+
+```
+
 ## C
+
 二叉树最大深度递归
 ```c
 int maxDepth(struct TreeNode* root){

problems/0110.平衡二叉树.md

@@ -670,7 +670,26 @@ var isBalanced = function (root) {
 };
 ```
 
+## TypeScript
+
+```typescript
+// 递归法
+function isBalanced(root: TreeNode | null): boolean {
+    function getDepth(root: TreeNode | null): number {
+        if (root === null) return 0;
+        let leftDepth: number = getDepth(root.left);
+        if (leftDepth === -1) return -1;
+        let rightDepth: number = getDepth(root.right);
+        if (rightDepth === -1) return -1;
+        if (Math.abs(leftDepth - rightDepth) > 1) return -1;
+        return 1 + Math.max(leftDepth, rightDepth);
+    }
+    return getDepth(root) !== -1;
+};
+```
+
 ## C
+
 递归法：
 ```c
 int getDepth(struct TreeNode* node) {

problems/0111.二叉树的最小深度.md

@@ -404,6 +404,44 @@ var minDepth = function(root) {
 };
 ```
 
+## TypeScript
+
+> 递归法
+
+```typescript
+function minDepth(root: TreeNode | null): number {
+    if (root === null) return 0;
+    if (root.left !== null && root.right === null) {
+        return 1 + minDepth(root.left);
+    }
+    if (root.left === null && root.right !== null) {
+        return 1 + minDepth(root.right);
+    }
+    return 1 + Math.min(minDepth(root.left), minDepth(root.right));
+}
+```
+
+> 迭代法
+
+```typescript
+function minDepth(root: TreeNode | null): number {
+    let helperQueue: TreeNode[] = [];
+    let resMin: number = 0;
+    let tempNode: TreeNode;
+    if (root !== null) helperQueue.push(root);
+    while (helperQueue.length > 0) {
+        resMin++;
+        for (let i = 0, length = helperQueue.length; i < length; i++) {
+            tempNode = helperQueue.shift()!;
+            if (tempNode.left === null && tempNode.right === null) return resMin;
+            if (tempNode.left !== null) helperQueue.push(tempNode.left);
+            if (tempNode.right !== null) helperQueue.push(tempNode.right);
+        }
+    }
+    return resMin;
+};
+```
+
 ## Swift
 
 > 递归

problems/0112.路径总和.md

@@ -531,82 +531,63 @@ class solution:
 ```go
 //递归法
 /**
- * definition for a binary tree node.
+ * Definition for a binary tree node.
- * type treenode struct {
+ * type TreeNode struct {
- *     val int
+ *     Val int
- *     left *treenode
+ *     Left *TreeNode
- *     right *treenode
+ *     Right *TreeNode
  * }
  */
-func haspathsum(root *treenode, targetsum int) bool {
+func hasPathSum(root *TreeNode, targetSum int) bool {
-    var flage bool //找没找到的标志
+    if root == nil {
-    if root==nil{
+        return false
-        return flage
     }
-    pathsum(root,0,targetsum,&flage)
+    
-    return flage
+    targetSum -= root.Val // 将targetSum在遍历每层的时候都减去本层节点的值
-}
+    if root.Left == nil && root.Right == nil && targetSum == 0 { // 如果剩余的targetSum为0, 则正好就是符合的结果
-func pathsum(root *treenode, sum int,targetsum int,flage *bool){
+        return true
-    sum+=root.val
-    if root.left==nil&&root.right==nil&&sum==targetsum{
-        *flage=true
-        return
-    }
-    if root.left!=nil&&!(*flage){//左节点不为空且还没找到
-        pathsum(root.left,sum,targetsum,flage)
-    } 
-    if root.right!=nil&&!(*flage){//右节点不为空且没找到
-        pathsum(root.right,sum,targetsum,flage)
     }
+    return hasPathSum(root.Left, targetSum) || hasPathSum(root.Right, targetSum) // 否则递归找
 }
 ```
 
-113 递归法
+113. 路径总和 II
 
 ```go
 /**
- * definition for a binary tree node.
+ * Definition for a binary tree node.
- * type treenode struct {
+ * type TreeNode struct {
- *     val int
+ *     Val int
- *     left *treenode
+ *     Left *TreeNode
- *     right *treenode
+ *     Right *TreeNode
  * }
  */
-func pathsum(root *treenode, targetsum int) [][]int {
+func pathSum(root *TreeNode, targetSum int) [][]int {
-    var result [][]int//最终结果
+    result := make([][]int, 0)
-    if root==nil{
+    traverse(root, &result, new([]int), targetSum)
-        return result
-    }
-    var sumnodes []int//经过路径的节点集合
-    haspathsum(root,&sumnodes,targetsum,&result)
     return result
 }
-func haspathsum(root *treenode,sumnodes *[]int,targetsum int,result *[][]int){
+
-    *sumnodes=append(*sumnodes,root.val)
+func traverse(node *TreeNode, result *[][]int, currPath *[]int, targetSum int) {
-    if root.left==nil&&root.right==nil{//叶子节点
+    if node == nil { // 这个判空也可以挪到递归遍历左右子树时去判断
-        fmt.println(*sumnodes)
+        return
-        var sum int
-        var number int
-        for k,v:=range *sumnodes{//求该路径节点的和
-            sum+=v
-            number=k
     }
-        tempnodes:=make([]int,number+1)//新的nodes接受指针里的值，防止最终指针里的值发生变动，导致最后的结果都是最后一个sumnodes的值
+
-        for k,v:=range *sumnodes{
+    targetSum -= node.Val // 将targetSum在遍历每层的时候都减去本层节点的值
-            tempnodes[k]=v
+    *currPath = append(*currPath, node.Val) // 把当前节点放到路径记录里
+    
+    if node.Left == nil && node.Right == nil && targetSum == 0 { // 如果剩余的targetSum为0, 则正好就是符合的结果
+        // 不能直接将currPath放到result里面, 因为currPath是共享的, 每次遍历子树时都会被修改
+        pathCopy := make([]int, len(*currPath)) 
+        for i, element := range *currPath {
+            pathCopy[i] = element
         }
-        if sum==targetsum{
+        *result = append(*result, pathCopy) // 将副本放到结果集里
-            *result=append(*result,tempnodes)
-        }
-    }
-    if root.left!=nil{
-        haspathsum(root.left,sumnodes,targetsum,result)
-        *sumnodes=(*sumnodes)[:len(*sumnodes)-1]//回溯
-    }
-    if root.right!=nil{
-        haspathsum(root.right,sumnodes,targetsum,result)
-        *sumnodes=(*sumnodes)[:len(*sumnodes)-1]//回溯
     }
+
+    traverse(node.Left, result, currPath, targetSum)
+    traverse(node.Right, result, currPath, targetSum)
+    *currPath = (*currPath)[:len(*currPath)-1] // 当前节点遍历完成, 从路径记录里删除掉
 }
 ```
 

problems/0139.单词拆分.md

@@ -248,6 +248,36 @@ class Solution {
         return valid[s.length()];
     }
 }
+
+// 回溯法+记忆化
+class Solution {
+    public boolean wordBreak(String s, List<String> wordDict) {
+        Set<String> wordDictSet = new HashSet(wordDict);
+        int[] memory = new int[s.length()];
+        return backTrack(s, wordDictSet, 0, memory);
+    }
+    
+    public boolean backTrack(String s, Set<String> wordDictSet, int startIndex, int[] memory) {
+        // 结束条件
+        if (startIndex >= s.length()) {
+            return true;
+        }
+        if (memory[startIndex] != 0) {
+            // 此处认为：memory[i] = 1 表示可以拼出i 及以后的字符子串， memory[i] = -1 表示不能
+            return memory[startIndex] == 1 ? true : false;
+        }
+        for (int i = startIndex; i < s.length(); ++i) {
+            // 处理 递归 回溯       循环不变量：[startIndex, i + 1)
+            String word = s.substring(startIndex, i + 1);
+            if (wordDictSet.contains(word) && backTrack(s, wordDictSet, i + 1, memory)) {
+                memory[startIndex] = 1;
+                return true;
+            }
+        }
+        memory[startIndex] = -1;
+        return false;
+    }
+}
 ```
 
 Python：

problems/0150.逆波兰表达式求值.md

@@ -210,6 +210,71 @@ var evalRPN = function(tokens) {
 };
 ```
 
+TypeScript：
+
+普通版：
+
+```typescript
+function evalRPN(tokens: string[]): number {
+    let helperStack: number[] = [];
+    let temp: number;
+    let i: number = 0;
+    while (i < tokens.length) {
+        let t: string = tokens[i];
+        switch (t) {
+            case '+':
+                temp = helperStack.pop()! + helperStack.pop()!;
+                helperStack.push(temp);
+                break;
+            case '-':
+                temp = helperStack.pop()!;
+                temp = helperStack.pop()! - temp;
+                helperStack.push(temp);
+                break;
+            case '*':
+                temp = helperStack.pop()! * helperStack.pop()!;
+                helperStack.push(temp);
+                break;
+            case '/':
+                temp = helperStack.pop()!;
+                temp = Math.trunc(helperStack.pop()! / temp);
+                helperStack.push(temp);
+                break;
+            default:
+                helperStack.push(Number(t));
+                break;
+        }
+        i++;
+    }
+    return helperStack.pop()!;
+};
+```
+
+优化版：
+
+```typescript
+function evalRPN(tokens: string[]): number {
+    const helperStack: number[] = [];
+    const operatorMap: Map<string, (a: number, b: number) => number> = new Map([
+        ['+', (a, b) => a + b],
+        ['-', (a, b) => a - b],
+        ['/', (a, b) => Math.trunc(a / b)],
+        ['*', (a, b) => a * b],
+    ]);
+    let a: number, b: number;
+    for (let t of tokens) {
+        if (operatorMap.has(t)) {
+            b = helperStack.pop()!;
+            a = helperStack.pop()!;
+            helperStack.push(operatorMap.get(t)!(a, b));
+        } else {
+            helperStack.push(Number(t));
+        }
+    }
+    return helperStack.pop()!;
+};
+```
+
 python3
 
 ```python

problems/0222.完全二叉树的节点个数.md

@@ -447,7 +447,63 @@ var countNodes = function(root) {
 };
 ```
 
+## TypeScrpt:
+
+> 递归法
+
+```typescript
+function countNodes(root: TreeNode | null): number {
+    if (root === null) return 0;
+    return 1 + countNodes(root.left) + countNodes(root.right);
+};
+```
+
+> 迭代法
+
+```typescript
+function countNodes(root: TreeNode | null): number {
+    let helperQueue: TreeNode[] = [];
+    let resCount: number = 0;
+    let tempNode: TreeNode;
+    if (root !== null) helperQueue.push(root);
+    while (helperQueue.length > 0) {
+        for (let i = 0, length = helperQueue.length; i < length; i++) {
+            tempNode = helperQueue.shift()!;
+            resCount++;
+            if (tempNode.left) helperQueue.push(tempNode.left);
+            if (tempNode.right) helperQueue.push(tempNode.right);
+        }
+    }
+    return resCount;
+};
+```
+
+> 利用完全二叉树性质
+
+```typescript
+function countNodes(root: TreeNode | null): number {
+    if (root === null) return 0;
+    let left: number = 0,
+        right: number = 0;
+    let curNode: TreeNode | null= root;
+    while (curNode !== null) {
+        left++;
+        curNode = curNode.left;
+    }
+    curNode = root;
+    while (curNode !== null) {
+        right++;
+        curNode = curNode.right;
+    }
+    if (left === right) {
+        return 2 ** left - 1;
+    }
+    return 1 + countNodes(root.left) + countNodes(root.right);
+};
+```
+
 ## C:
+
 递归法
 ```c
 int countNodes(struct TreeNode* root) {

problems/0226.翻转二叉树.md

@@ -563,7 +563,135 @@ var invertTree = function(root) {
 };
 ```
 
+### TypeScript：
+
+递归法：
+
+```typescript
+// 递归法（前序遍历）
+function invertTree(root: TreeNode | null): TreeNode | null {
+    if (root === null) return root;
+    let tempNode: TreeNode | null = root.left;
+    root.left = root.right;
+    root.right = tempNode;
+    invertTree(root.left);
+    invertTree(root.right);
+    return root;
+};
+
+// 递归法（后序遍历）
+function invertTree(root: TreeNode | null): TreeNode | null {
+    if (root === null) return root;
+    invertTree(root.left);
+    invertTree(root.right);
+    let tempNode: TreeNode | null = root.left;
+    root.left = root.right;
+    root.right = tempNode;
+    return root;
+};
+
+// 递归法（中序遍历）
+function invertTree(root: TreeNode | null): TreeNode | null {
+    if (root === null) return root;
+    invertTree(root.left);
+    let tempNode: TreeNode | null = root.left;
+    root.left = root.right;
+    root.right = tempNode;
+    // 因为左右节点已经进行交换，此时的root.left 是原先的root.right
+    invertTree(root.left);
+    return root;
+};
+```
+
+迭代法：
+
+```typescript
+// 迭代法（栈模拟前序遍历）
+function invertTree(root: TreeNode | null): TreeNode | null {
+    let helperStack: TreeNode[] = [];
+    let curNode: TreeNode,
+        tempNode: TreeNode | null;
+    if (root !== null) helperStack.push(root);
+    while (helperStack.length > 0) {
+        curNode = helperStack.pop()!;
+        // 入栈操作最好在交换节点之前进行，便于理解
+        if (curNode.right) helperStack.push(curNode.right);
+        if (curNode.left) helperStack.push(curNode.left);
+        tempNode = curNode.left;
+        curNode.left = curNode.right;
+        curNode.right = tempNode;
+    }
+    return root;
+};
+
+// 迭代法（栈模拟中序遍历-统一写法形式）
+function invertTree(root: TreeNode | null): TreeNode | null {
+    let helperStack: (TreeNode | null)[] = [];
+    let curNode: TreeNode | null,
+        tempNode: TreeNode | null;
+    if (root !== null) helperStack.push(root);
+    while (helperStack.length > 0) {
+        curNode = helperStack.pop();
+        if (curNode !== null) {
+            if (curNode.right !== null) helperStack.push(curNode.right);
+            helperStack.push(curNode);
+            helperStack.push(null);
+            if (curNode.left !== null) helperStack.push(curNode.left);
+        } else {
+            curNode = helperStack.pop()!;
+            tempNode = curNode.left;
+            curNode.left = curNode.right;
+            curNode.right = tempNode;
+        }
+    }
+    return root;
+};
+
+// 迭代法（栈模拟后序遍历-统一写法形式）
+function invertTree(root: TreeNode | null): TreeNode | null {
+    let helperStack: (TreeNode | null)[] = [];
+    let curNode: TreeNode | null,
+        tempNode: TreeNode | null;
+    if (root !== null) helperStack.push(root);
+    while (helperStack.length > 0) {
+        curNode = helperStack.pop();
+        if (curNode !== null) {
+            helperStack.push(curNode);
+            helperStack.push(null);
+            if (curNode.right !== null) helperStack.push(curNode.right);
+            if (curNode.left !== null) helperStack.push(curNode.left);
+        } else {
+            curNode = helperStack.pop()!;
+            tempNode = curNode.left;
+            curNode.left = curNode.right;
+            curNode.right = tempNode;
+        }
+    }
+    return root;
+};
+
+// 迭代法（队列模拟层序遍历）
+function invertTree(root: TreeNode | null): TreeNode | null {
+    const helperQueue: TreeNode[] = [];
+    let curNode: TreeNode,
+        tempNode: TreeNode | null;
+    if (root !== null) helperQueue.push(root);
+    while (helperQueue.length > 0) {
+        for (let i = 0, length = helperQueue.length; i < length; i++) {
+            curNode = helperQueue.shift()!;
+            tempNode = curNode.left;
+            curNode.left = curNode.right;
+            curNode.right = tempNode;
+            if (curNode.left !== null) helperQueue.push(curNode.left);
+            if (curNode.right !== null) helperQueue.push(curNode.right);
+        }
+    }
+    return root;
+};
+```
+
 ### C:
+
 递归法
 ```c
 struct TreeNode* invertTree(struct TreeNode* root){

problems/0232.用栈实现队列.md

@@ -348,7 +348,44 @@ MyQueue.prototype.empty = function() {
 };
  ```
 
+TypeScript:
+
+```typescript
+class MyQueue {
+    private stackIn: number[]
+    private stackOut: number[]
+    constructor() {
+        this.stackIn = [];
+        this.stackOut = [];
+    }
+
+    push(x: number): void {
+        this.stackIn.push(x);
+    }
+
+    pop(): number {
+        if (this.stackOut.length === 0) {
+            while (this.stackIn.length > 0) {
+                this.stackOut.push(this.stackIn.pop()!);
+            }
+        }
+        return this.stackOut.pop()!;
+    }
+
+    peek(): number {
+        let temp: number = this.pop();
+        this.stackOut.push(temp);
+        return temp;
+    }
+
+    empty(): boolean {
+        return this.stackIn.length === 0 && this.stackOut.length === 0;
+    }
+}
+```
+
 Swift：
+
 ```swift
 class MyQueue {
     

problems/0239.滑动窗口最大值.md

@@ -395,30 +395,102 @@ func maxSlidingWindow(nums []int, k int) []int {
 
 Javascript:
 ```javascript
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number[]}
+ */
 var maxSlidingWindow = function (nums, k) {
-  // 队列数组（存放的是元素下标，为了取值方便）
+    class MonoQueue {
-  const q = [];
+        queue;
-  // 结果数组
+        constructor() {
-  const ans = [];
+            this.queue = [];
-  for (let i = 0; i < nums.length; i++) {
-    // 若队列不为空，且当前元素大于等于队尾所存下标的元素，则弹出队尾
-    while (q.length && nums[i] >= nums[q[q.length - 1]]) {
-      q.pop();
         }
-    // 入队当前元素下标
+        enqueue(value) {
-    q.push(i);
+            let back = this.queue[this.queue.length - 1];
-    // 判断当前最大值（即队首元素）是否在窗口中，若不在便将其出队
+            while (back !== undefined && back < value) {
-    if (q[0] <= i - k) {
+                this.queue.pop();
-      q.shift();
+                back = this.queue[this.queue.length - 1];
             }
-    // 当达到窗口大小时便开始向结果中添加数据
+            this.queue.push(value);
-    if (i >= k - 1) ans.push(nums[q[0]]);
         }
-  return ans;
+        dequeue(value) {
+            let front = this.front();
+            if (front === value) {
+                this.queue.shift();
+            }
+        }
+        front() {
+            return this.queue[0];
+        }
+    }
+    let helperQueue = new MonoQueue();
+    let i = 0, j = 0;
+    let resArr = [];
+    while (j < k) {
+        helperQueue.enqueue(nums[j++]);
+    }
+    resArr.push(helperQueue.front());
+    while (j < nums.length) {
+        helperQueue.enqueue(nums[j]);
+        helperQueue.dequeue(nums[i]);
+        resArr.push(helperQueue.front());
+        i++, j++;
+    }
+    return resArr;
+};
+```
+
+TypeScript：
+
+```typescript
+function maxSlidingWindow(nums: number[], k: number): number[] {
+    /** 单调递减队列 */
+    class MonoQueue {
+        private queue: number[];
+        constructor() {
+            this.queue = [];
+        };
+        /** 入队：value如果大于队尾元素，则将队尾元素删除，直至队尾元素大于value，或者队列为空 */
+        public enqueue(value: number): void {
+            let back: number | undefined = this.queue[this.queue.length - 1];
+            while (back !== undefined && back < value) {
+                this.queue.pop();
+                back = this.queue[this.queue.length - 1];
+            }
+            this.queue.push(value);
+        };
+        /** 出队：只有当队头元素等于value，才出队 */
+        public dequeue(value: number): void {
+            let top: number | undefined = this.top();
+            if (top !== undefined && top === value) {
+                this.queue.shift();
+            }
+        }
+        public top(): number | undefined {
+            return this.queue[0];
+        }
+    }
+    const helperQueue: MonoQueue = new MonoQueue();
+    let i: number = 0,
+        j: number = 0;
+    let resArr: number[] = [];
+    while (j < k) {
+        helperQueue.enqueue(nums[j++]);
+    }
+    resArr.push(helperQueue.top()!);
+    while (j < nums.length) {
+        helperQueue.enqueue(nums[j]);
+        helperQueue.dequeue(nums[i]);
+        resArr.push(helperQueue.top()!);
+        j++, i++;
+    }
+    return resArr;
 };
 ```
 
 Swift:
+
 ```Swift
 /// 双向链表
 class DoublyListNode {

problems/0343.整数拆分.md

@@ -4,23 +4,22 @@
 </a>
 <p align="center"><strong><a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
 
-## 343. 整数拆分
+# 343. 整数拆分
 
 [力扣题目链接](https://leetcode-cn.com/problems/integer-break/)
 
 给定一个正整数 n，将其拆分为至少两个正整数的和，并使这些整数的乘积最大化。 返回你可以获得的最大乘积。
 
 示例 1:
-输入: 2
+* 输入: 2
-输出: 1
+* 输出: 1
-
+* 解释: 2 = 1 + 1, 1 × 1 = 1。
-\解释: 2 = 1 + 1, 1 × 1 = 1。
 
 示例 2:
-输入: 10
+* 输入: 10
-输出: 36
+* 输出: 36
-解释: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36。
+* 解释: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36。
-说明: 你可以假设 n 不小于 2 且不大于 58。
+* 说明: 你可以假设 n 不小于 2 且不大于 58。
 
 ## 思路
 
@@ -193,7 +192,7 @@ public:
 ## 其他语言版本
 
 
-Java：
+### Java 
 ```Java
 class Solution {
     public int integerBreak(int n) {
@@ -212,7 +211,7 @@ class Solution {
 }
 ```
 
-Python：
+### Python
 ```python
 class Solution:
     def integerBreak(self, n: int) -> int:
@@ -226,7 +225,8 @@ class Solution:
                 dp[i] = max(dp[i], max(j * (i - j), j * dp[i - j]))
         return dp[n]
 ```
-Go：
+
+### Go
 ```golang
 func integerBreak(n int) int {
     /**
@@ -256,7 +256,7 @@ func max(a,b int) int{
 }
 ```
 
-Javascript:
+### Javascript
 ```Javascript
 var integerBreak = function(n) {
     let dp = new Array(n + 1).fill(0)

problems/0347.前K个高频元素.md

@@ -358,6 +358,22 @@ PriorityQueue.prototype.compare = function(index1, index2) {
 }
 ```
 
+TypeScript：
+
+```typescript
+function topKFrequent(nums: number[], k: number): number[] {
+    const countMap: Map<number, number> = new Map();
+    for (let num of nums) {
+        countMap.set(num, (countMap.get(num) || 0) + 1);
+    }
+    // tS没有最小堆的数据结构，所以直接对整个数组进行排序，取前k个元素
+    return [...countMap.entries()]
+        .sort((a, b) => b[1] - a[1])
+        .slice(0, k)
+        .map(i => i[0]);
+};
+```
+
 
 
 -----------------------

problems/0376.摆动序列.md

@@ -174,7 +174,7 @@ public:
 ```Java
 class Solution {
     public int wiggleMaxLength(int[] nums) {
-        if (nums == null || nums.length <= 1) {
+        if (nums.length <= 1) {
             return nums.length;
         }
         //当前差值

problems/0416.分割等和子集.md

@@ -210,6 +210,45 @@ class Solution {
 }
 ```
 
+二维数组版本（易于理解）：
+```Java
+class Solution {
+    public boolean canPartition(int[] nums) {
+        int sum = 0;
+        for (int i = 0; i < nums.length; i++) {
+            sum += nums[i];
+        }
+
+        if (sum % 2 == 1)
+            return false;
+        int target = sum / 2;
+
+        //dp[i][j]代表可装物品为0-i，背包容量为j的情况下，背包内容量的最大价值
+        int[][] dp = new int[nums.length][target + 1];
+
+        //初始化,dp[0][j]的最大价值nums[0](if j > weight[i])
+        //dp[i][0]均为0，不用初始化
+        for (int j = nums[0]; j <= target; j++) {
+            dp[0][j] = nums[0];
+        }
+
+        //遍历物品，遍历背包
+        //递推公式:
+        for (int i = 1; i < nums.length; i++) {
+            for (int j = 0; j <= target; j++) {
+                //背包容量可以容纳nums[i]
+                if (j >= nums[i]) {
+                    dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - nums[i]] + nums[i]);
+                } else {
+                    dp[i][j] = dp[i - 1][j];
+                }
+            }
+        }
+
+        return dp[nums.length - 1][target] == target;
+    }
+}
+```
 Python：
 ```python
 class Solution:

problems/0435.无重叠区间.md

@@ -183,28 +183,22 @@ public:
 ```java
 class Solution {
     public int eraseOverlapIntervals(int[][] intervals) {
-        if (intervals.length < 2) return 0;
+        Arrays.sort(intervals, (a, b) -> {
-
+            if (a[0] == a[0]) return a[1] - b[1];
-        Arrays.sort(intervals, new Comparator<int[]>() {
+            return a[0] - b[0];
-            @Override
-            public int compare(int[] o1, int[] o2) {
-                if (o1[1] != o2[1]) {
-                    return Integer.compare(o1[1],o2[1]);
-                } else {
-                    return Integer.compare(o1[0],o2[0]);
-                }
-            }
         });
 
-        int count = 1;
+        int count = 0;
-        int edge = intervals[0][1];
+        int edge = Integer.MIN_VALUE;
-        for (int i = 1; i < intervals.length; i++) {
+        for (int i = 0; i < intervals.length; i++) {
-            if (edge <= intervals[i][0]){
+            if (edge <= intervals[i][0]) {
-                count ++; //non overlap + 1
                 edge = intervals[i][1];
+            } else {
+                count++;
             }
         }
-        return intervals.length - count;
+
+        return count;
     }
 }
 ```

problems/0494.目标和.md

@@ -272,7 +272,8 @@ Python：
 class Solution:
     def findTargetSumWays(self, nums: List[int], target: int) -> int:
         sumValue = sum(nums)
-        if target > sumValue or (sumValue + target) % 2 == 1: return 0
+        #注意边界条件为 target>sumValue or target<-sumValue or  (sumValue + target) % 2 == 1
+        if abs(target) > sumValue or (sumValue + target) % 2 == 1: return 0
         bagSize = (sumValue + target) // 2
         dp = [0] * (bagSize + 1)
         dp[0] = 1

problems/1047.删除字符串中的所有相邻重复项.md

@@ -267,8 +267,32 @@ var removeDuplicates = function(s) {
 };
 ```
 
+TypeScript：
+
+```typescript
+function removeDuplicates(s: string): string {
+    const helperStack: string[] = [];
+    let i: number = 0;
+    while (i < s.length) {
+        let top: string = helperStack[helperStack.length - 1];
+        if (top === s[i]) {
+            helperStack.pop();
+        } else {
+            helperStack.push(s[i]);
+        }
+        i++;
+    }
+    let res: string = '';
+    while (helperStack.length > 0) {
+        res = helperStack.pop() + res;
+    }
+    return res;
+};
+```
+
 C:
 方法一：使用栈
+
 ```c
 char * removeDuplicates(char * s){
     //求出字符串长度

problems/1049.最后一块石头的重量II.md

@@ -153,6 +153,8 @@ public:
 
 
 Java：
+
+一维数组版本
 ```Java
 class Solution {
     public int lastStoneWeightII(int[] stones) {
@@ -173,6 +175,41 @@ class Solution {
         return sum - 2 * dp[target];
     }
 }
+```
+二维数组版本（便于理解）
+```Java
+class Solution {
+    public int lastStoneWeightII(int[] stones) {
+        int sum = 0;
+        for (int s : stones) {
+            sum += s;
+        }
+
+        int target = sum / 2;
+        //初始化，dp[i][j]为可以放0-i物品，背包容量为j的情况下背包中的最大价值
+        int[][] dp = new int[stones.length][target + 1];
+        //dp[i][0]默认初始化为0
+        //dp[0][j]取决于stones[0]
+        for (int j = stones[0]; j <= target; j++) {
+            dp[0][j] = stones[0];
+        }
+
+        for (int i = 1; i < stones.length; i++) {
+            for (int j = 1; j <= target; j++) {//注意是等于
+                if (j >= stones[i]) {
+                    //不放:dp[i - 1][j] 放:dp[i - 1][j - stones[i]] + stones[i]
+                    dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - stones[i]] + stones[i]);
+                } else {
+                    dp[i][j] = dp[i - 1][j];
+                }
+            }
+        }
+
+        System.out.println(dp[stones.length - 1][target]);
+        return (sum - dp[stones.length - 1][target]) - dp[stones.length - 1][target];
+    }
+}
+
 ```
 
 Python：

problems/二叉树理论基础.md

@@ -227,7 +227,23 @@ function TreeNode(val, left, right) {
 }
 ```
 
+TypeScript：
+
+```typescript
+class TreeNode {
+    public val: number;
+    public left: TreeNode | null;
+    public right: TreeNode | null;
+    constructor(val?: number, left?: TreeNode, right?: TreeNode) {
+        this.val = val === undefined ? 0 : val;
+        this.left = left === undefined ? null : left;
+        this.right = right === undefined ? null : right;
+    }
+}
+```
+
 Swift:
+
 ```Swift
 class TreeNode<T> {
     var value: T

problems/二叉树的统一迭代法.md

@@ -522,7 +522,75 @@ var postorderTraversal = function(root, res = []) {
 
 ```
 
+TypeScript：
 
+```typescript
+// 前序遍历（迭代法）
+function preorderTraversal(root: TreeNode | null): number[] {
+    let helperStack: (TreeNode | null)[] = [];
+    let res: number[] = [];
+    let curNode: TreeNode | null;
+    if (root === null) return res;
+    helperStack.push(root);
+    while (helperStack.length > 0) {
+        curNode = helperStack.pop()!;
+        if (curNode !== null) {
+            if (curNode.right !== null) helperStack.push(curNode.right);
+            helperStack.push(curNode);
+            helperStack.push(null);
+            if (curNode.left !== null) helperStack.push(curNode.left);
+        } else {
+            curNode = helperStack.pop()!;
+            res.push(curNode.val);
+        }
+    }
+    return res;
+};
+
+// 中序遍历（迭代法）
+function inorderTraversal(root: TreeNode | null): number[] {
+    let helperStack: (TreeNode | null)[] = [];
+    let res: number[] = [];
+    let curNode: TreeNode | null;
+    if (root === null) return res;
+    helperStack.push(root);
+    while (helperStack.length > 0) {
+        curNode = helperStack.pop()!;
+        if (curNode !== null) {
+            if (curNode.right !== null) helperStack.push(curNode.right);
+            helperStack.push(curNode);
+            helperStack.push(null);
+            if (curNode.left !== null) helperStack.push(curNode.left);
+        } else {
+            curNode = helperStack.pop()!;
+            res.push(curNode.val);
+        }
+    }
+    return res;
+};
+
+// 后序遍历（迭代法）
+function postorderTraversal(root: TreeNode | null): number[] {
+    let helperStack: (TreeNode | null)[] = [];
+    let res: number[] = [];
+    let curNode: TreeNode | null;
+    if (root === null) return res;
+    helperStack.push(root);
+    while (helperStack.length > 0) {
+        curNode = helperStack.pop()!;
+        if (curNode !== null) {
+            if (curNode.right !== null) helperStack.push(curNode.right);
+            helperStack.push(curNode);
+            helperStack.push(null);
+            if (curNode.left !== null) helperStack.push(curNode.left);
+        } else {
+            curNode = helperStack.pop()!;
+            res.push(curNode.val);
+        }
+    }
+    return res;
+};
+```
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/二叉树的迭代遍历.md

@@ -454,6 +454,61 @@ var postorderTraversal = function(root, res = []) {
 };
 ```
 
+TypeScript:
+
+```typescript
+// 前序遍历（迭代法）
+function preorderTraversal(root: TreeNode | null): number[] {
+    if (root === null) return [];
+    let res: number[] = [];
+    let helperStack: TreeNode[] = [];
+    let curNode: TreeNode = root;
+    helperStack.push(curNode);
+    while (helperStack.length > 0) {
+        curNode = helperStack.pop()!;
+        res.push(curNode.val);
+        if (curNode.right !== null) helperStack.push(curNode.right);
+        if (curNode.left !== null) helperStack.push(curNode.left);
+    }
+    return res;
+};
+
+// 中序遍历（迭代法）
+function inorderTraversal(root: TreeNode | null): number[] {
+    let helperStack: TreeNode[] = [];
+    let res: number[] = [];
+    if (root === null) return res;
+    let curNode: TreeNode | null = root;
+    while (curNode !== null || helperStack.length > 0) {
+        if (curNode !== null) {
+            helperStack.push(curNode);
+            curNode = curNode.left;
+        } else {
+            curNode = helperStack.pop()!;
+            res.push(curNode.val);
+            curNode = curNode.right;
+        }
+    }
+    return res;
+};
+
+// 后序遍历（迭代法）
+function postorderTraversal(root: TreeNode | null): number[] {
+    let helperStack: TreeNode[] = [];
+    let res: number[] = [];
+    let curNode: TreeNode;
+    if (root === null) return res;
+    helperStack.push(root);
+    while (helperStack.length > 0) {
+        curNode = helperStack.pop()!;
+        res.push(curNode.val);
+        if (curNode.left !== null) helperStack.push(curNode.left);
+        if (curNode.right !== null) helperStack.push(curNode.right);
+    }
+    return res.reverse();
+};
+```
+
 Swift:
 
 > 迭代法前序遍历

problems/二叉树的递归遍历.md

@@ -270,40 +270,6 @@ func postorderTraversal(root *TreeNode) (res []int) {
 }
 ```
 
-javaScript:
-
-```js
-
-前序遍历:
-
-var preorderTraversal = function(root, res = []) {
-    if (!root) return res;
-    res.push(root.val);
-    preorderTraversal(root.left, res)
-    preorderTraversal(root.right, res)
-    return res;
-};
-
-中序遍历:
-
-var inorderTraversal = function(root, res = []) {
-    if (!root) return res;
-    inorderTraversal(root.left, res);
-    res.push(root.val);
-    inorderTraversal(root.right, res);
-    return res;
-};
-
-后序遍历:
-
-var postorderTraversal = function(root, res = []) {
-    if (!root) return res;
-    postorderTraversal(root.left, res);
-    postorderTraversal(root.right, res);
-    res.push(root.val);
-    return res;
-};
-```
 Javascript版本：
 
 前序遍历：
@@ -358,7 +324,51 @@ var postorderTraversal = function(root) {
 };
 ```
 
+TypeScript:
+
+```typescript
+// 前序遍历
+function preorderTraversal(node: TreeNode | null): number[] {
+    function traverse(node: TreeNode | null, res: number[]): void {
+        if (node === null) return;
+        res.push(node.val);
+        traverse(node.left, res);
+        traverse(node.right, res);
+    }
+    const res: number[] = [];
+    traverse(node, res);
+    return res;
+}
+
+// 中序遍历
+function inorderTraversal(node: TreeNode | null): number[] {
+    function traverse(node: TreeNode | null, res: number[]): void {
+        if (node === null) return;
+        traverse(node.left, res);
+        res.push(node.val);
+        traverse(node.right, res);
+    }
+    const res: number[] = [];
+    traverse(node, res);
+    return res;
+}
+
+// 后序遍历
+function postorderTraversal(node: TreeNode | null): number[] {
+    function traverse(node: TreeNode | null, res: number[]): void {
+        if (node === null) return;
+        traverse(node.left, res);
+        traverse(node.right, res);
+        res.push(node.val);
+    }
+    const res: number[] = [];
+    traverse(node, res);
+    return res;
+}
+```
+
 C:
+
 ```c
 //前序遍历：
 void preOrderTraversal(struct TreeNode* root, int* ret, int* returnSize) {

problems/前序/程序员写文档工具.md

@@ -135,7 +135,6 @@ Markdown支持部分html，例如这样
 
 
 
-
 -----------------------
 * 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/背包理论基础01背包-1.md

@@ -3,11 +3,12 @@
   <img src="https://code-thinking-1253855093.file.myqcloud.com/pics/20210924105952.png" width="1000"/>
 </a>
 <p align="center"><strong><a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
+
 # 动态规划：关于01背包问题，你该了解这些！
 
 这周我们正式开始讲解背包问题！
 
-背包问题的经典资料当然是：背包九讲。在公众号「代码随想录」后台回复：背包九讲，就可以获得背包九讲的PDF。
+背包问题的经典资料当然是：背包九讲。在公众号「代码随想录」后台回复：背包九讲，就可以获得背包九讲的pdf。
 
 但说实话，背包九讲对于小白来说确实不太友好，看起来还是有点费劲的，而且都是伪代码理解起来也吃力。
 
@@ -32,7 +33,7 @@ leetcode上没有纯01背包的问题，都是01背包应用方面的题目，
 
 ## 01 背包
 
-有N件物品和一个最多能背重量为W 的背包。第i件物品的重量是weight[i]，得到的价值是value[i] 。**每件物品只能用一次**，求解将哪些物品装入背包里物品价值总和最大。
+有n件物品和一个最多能背重量为w 的背包。第i件物品的重量是weight[i]，得到的价值是value[i] 。**每件物品只能用一次**，求解将哪些物品装入背包里物品价值总和最大。
 
 ![动态规划-背包问题](https://img-blog.csdnimg.cn/20210117175428387.jpg)
 
@@ -40,7 +41,7 @@ leetcode上没有纯01背包的问题，都是01背包应用方面的题目，
 
 这样其实是没有从底向上去思考，而是习惯性想到了背包，那么暴力的解法应该是怎么样的呢？
 
-每一件物品其实只有两个状态，取或者不取，所以可以使用回溯法搜索出所有的情况，那么时间复杂度就是$O(2^n)$，这里的n表示物品数量。
+每一件物品其实只有两个状态，取或者不取，所以可以使用回溯法搜索出所有的情况，那么时间复杂度就是$o(2^n)$，这里的n表示物品数量。
 
 **所以暴力的解法是指数级别的时间复杂度。进而才需要动态规划的解法来进行优化！**
 
@@ -109,7 +110,7 @@ for (int j = 0 ; j < weight[0]; j++) {  // 当然这一步，如果把dp数组
     dp[0][j] = 0;
 }
 // 正序遍历
-for (int j = weight[0]; j <= bagWeight; j++) {
+for (int j = weight[0]; j <= bagweight; j++) {
     dp[0][j] = value[0];
 }
 ```
@@ -135,8 +136,8 @@ dp[0][j] 和 dp[i][0] 都已经初始化了，那么其他下标应该初始化
 
 ```
 // 初始化 dp
-vector<vector<int>> dp(weight.size(), vector<int>(bagWeight + 1, 0));
+vector<vector<int>> dp(weight.size(), vector<int>(bagweight + 1, 0));
-for (int j = weight[0]; j <= bagWeight; j++) {
+for (int j = weight[0]; j <= bagweight; j++) {
     dp[0][j] = value[0];
 }
 
@@ -160,7 +161,7 @@ for (int j = weight[0]; j <= bagWeight; j++) {
 ```
 // weight数组的大小 就是物品个数
 for(int i = 1; i < weight.size(); i++) { // 遍历物品
-    for(int j = 0; j <= bagWeight; j++) { // 遍历背包容量
+    for(int j = 0; j <= bagweight; j++) { // 遍历背包容量
         if (j < weight[i]) dp[i][j] = dp[i - 1][j]; 
         else dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]);
 
@@ -174,7 +175,7 @@ for(int i = 1; i < weight.size(); i++) { // 遍历物品
 
 ```
 // weight数组的大小 就是物品个数
-for(int j = 0; j <= bagWeight; j++) { // 遍历背包容量
+for(int j = 0; j <= bagweight; j++) { // 遍历背包容量
     for(int i = 1; i < weight.size(); i++) { // 遍历物品
         if (j < weight[i]) dp[i][j] = dp[i - 1][j];
         else dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]);
@@ -219,32 +220,32 @@ dp[i-1][j]和dp[i - 1][j - weight[i]] 都在dp[i][j]的左上角方向（包括
 主要就是自己没有动手推导一下dp数组的演变过程，如果推导明白了，代码写出来就算有问题，只要把dp数组打印出来，对比一下和自己推导的有什么差异，很快就可以发现问题了。
 
 
-## 完整C++测试代码
+## 完整c++测试代码
 
-```CPP
+```cpp
 void test_2_wei_bag_problem1() {
     vector<int> weight = {1, 3, 4};
     vector<int> value = {15, 20, 30};
-    int bagWeight = 4;
+    int bagweight = 4;
 
     // 二维数组
-    vector<vector<int>> dp(weight.size(), vector<int>(bagWeight + 1, 0));
+    vector<vector<int>> dp(weight.size(), vector<int>(bagweight + 1, 0));
 
     // 初始化
-    for (int j = weight[0]; j <= bagWeight; j++) {
+    for (int j = weight[0]; j <= bagweight; j++) {
         dp[0][j] = value[0];
     }
 
     // weight数组的大小 就是物品个数
     for(int i = 1; i < weight.size(); i++) { // 遍历物品
-        for(int j = 0; j <= bagWeight; j++) { // 遍历背包容量
+        for(int j = 0; j <= bagweight; j++) { // 遍历背包容量
             if (j < weight[i]) dp[i][j] = dp[i - 1][j];
             else dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]);
 
         }
     }
 
-    cout << dp[weight.size() - 1][bagWeight] << endl;
+    cout << dp[weight.size() - 1][bagweight] << endl;
 }
 
 int main() {
@@ -267,48 +268,45 @@ int main() {
 
 ## 其他语言版本
 
-Java：
+### java 
 
 ```java
-    public static void main(String[] args) {
+    public static void main(string[] args) {
         int[] weight = {1, 3, 4};
         int[] value = {15, 20, 30};
-        int bagSize = 4;
+        int bagsize = 4;
-        testWeightBagProblem(weight, value, bagSize);
+        testweightbagproblem(weight, value, bagsize);
     }
 
-    public static void testWeightBagProblem(int[] weight, int[] value, int bagSize){
+    public static void testweightbagproblem(int[] weight, int[] value, int bagsize){
-        int wLen = weight.length, value0 = 0;
+        int wlen = weight.length, value0 = 0;
         //定义dp数组：dp[i][j]表示背包容量为j时，前i个物品能获得的最大价值
-        int[][] dp = new int[wLen + 1][bagSize + 1];
+        int[][] dp = new int[wlen + 1][bagsize + 1];
         //初始化：背包容量为0时，能获得的价值都为0
-        for (int i = 0; i <= wLen; i++){
+        for (int i = 0; i <= wlen; i++){
             dp[i][0] = value0;
         }
         //遍历顺序：先遍历物品，再遍历背包容量
-        for (int i = 1; i <= wLen; i++){
+        for (int i = 1; i <= wlen; i++){
-            for (int j = 1; j <= bagSize; j++){
+            for (int j = 1; j <= bagsize; j++){
                 if (j < weight[i - 1]){
                     dp[i][j] = dp[i - 1][j];
                 }else{
-                    dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - weight[i - 1]] + value[i - 1]);
+                    dp[i][j] = math.max(dp[i - 1][j], dp[i - 1][j - weight[i - 1]] + value[i - 1]);
                 }
             }
         }
         //打印dp数组
-        for (int i = 0; i <= wLen; i++){
+        for (int i = 0; i <= wlen; i++){
-            for (int j = 0; j <= bagSize; j++){
+            for (int j = 0; j <= bagsize; j++){
-                System.out.print(dp[i][j] + " ");
+                system.out.print(dp[i][j] + " ");
             }
-            System.out.print("\n");
+            system.out.print("\n");
         }
     }
 ```
 
-
+### python
-
-
-Python：
 ```python
 def test_2_wei_bag_problem1(bag_size, weight, value) -> int: 
 	rows, cols = len(weight), bag_size + 1
@@ -343,26 +341,26 @@ if __name__ == "__main__":
 ```
 
 
-Go：
+### go 
 ```go
-func test_2_wei_bag_problem1(weight, value []int, bagWeight int) int {
+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)
+		dp[i] = make([]int, bagweight+1)
 	}
 	// 初始化
-	for j := bagWeight; j >= weight[0]; j-- {
+	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++ {
+		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]
+	return dp[len(weight)-1][bagweight]
 }
 
 func max(a,b int) int {
@@ -379,19 +377,19 @@ func main() {
 }
 ```
 
-javaScript: 
+### javascript 
 
 ```js
-function testWeightBagProblem (wight, value, size) {
+function testweightbagproblem (wight, value, size) {
   const len = wight.length, 
-    dp = Array.from({length: len + 1}).map(
+    dp = array.from({length: len + 1}).map(
-      () => Array(size + 1).fill(0)
+      () => array(size + 1).fill(0)
     );
   
   for(let i = 1; i <= len; i++) {
     for(let j = 0; j <= size; j++) {
       if(wight[i - 1] <= j) {
-        dp[i][j] = Math.max(
+        dp[i][j] = math.max(
           dp[i - 1][j], 
           value[i - 1] + dp[i - 1][j - wight[i - 1]]
         )

problems/背包理论基础01背包-2.md

@@ -210,7 +210,7 @@ int main() {
 
 ## 其他语言版本
 
-Java：
+### Java 
 
 ```java
     public static void main(String[] args) {
@@ -240,7 +240,7 @@ Java：
 
 
 
-Python：
+### Python
 ```python
 def test_1_wei_bag_problem():
     weight = [1, 3, 4]
@@ -260,7 +260,7 @@ def test_1_wei_bag_problem():
 test_1_wei_bag_problem()
 ```
 
-Go：
+### Go 
 ```go
 func test_1_wei_bag_problem(weight, value []int, bagWeight int) int {
 	// 定义 and 初始化
@@ -292,7 +292,7 @@ func main() {
 }
 ```
 
-javaScript:
+### javaScript 
 
 ```js
 

problems/背包问题理论基础完全背包.md

@@ -183,13 +183,11 @@ private static void testCompletePack(){
     int[] value = {15, 20, 30};
     int bagWeight = 4;
     int[] dp = new int[bagWeight + 1];
-    for (int i = 0; i < weight.length; i++){
+    for (int i = 0; i < weight.length; i++){ // 遍历物品
-        for (int j = 1; j <= bagWeight; j++){
+        for (int j = weight[i]; j <= bagWeight; j++){ // 遍历背包容量
-            if (j - weight[i] >= 0){
             dp[j] = Math.max(dp[j], dp[j - weight[i]] + value[i]);
         }
     }
-    }
     for (int maxValue : dp){
         System.out.println(maxValue + "   ");
     }
@@ -201,8 +199,8 @@ private static void testCompletePackAnotherWay(){
     int[] value = {15, 20, 30};
     int bagWeight = 4;
     int[] dp = new int[bagWeight + 1];
-    for (int i = 1; i <= bagWeight; i++){
+    for (int i = 1; i <= bagWeight; i++){ // 遍历背包容量
-        for (int j = 0; j < weight.length; j++){
+        for (int j = 0; j < weight.length; j++){ // 遍历物品
             if (i - weight[j] >= 0){
                 dp[i] = Math.max(dp[i], dp[i - weight[j]] + value[j]);
             }