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

Author: youngyangyang04

problems/0001.两数之和.md

@@ -187,7 +187,23 @@ var twoSum = function (nums, target) {
 };
 ```
 
-
+php
+
+```php
+function twoSum(array $nums, int $target): array
+{
+    for ($i = 0; $i < count($nums);$i++) {
+        // 计算剩下的数
+        $residue = $target - $nums[$i];
+        // 匹配的index，有则返回index， 无则返回false
+        $match_index = array_search($residue, $nums);
+        if ($match_index !== false && $match_index != $i) {
+            return array($i, $match_index);
+        }
+    }
+    return [];
+}
+```
 
 
 -----------------------

problems/0015.三数之和.md

@@ -354,6 +354,47 @@ def is_valid(strs)
 end
 ```
 
+php:
+
+```php
+function threeSum(array $nums): array
+{
+    $result = [];
+    $length = count($nums);
+    if ($length < 3) {
+        return [];
+    }
+    sort($nums);
+    for ($i = 0; $i < $length; $i++) {
+        // 如果大于0结束
+        if ($nums[$i] > 0) break;
+        // 去重
+        if ($i > 0 && $nums[$i] == $nums[$i - 1]) continue;
+        $left = $i + 1;
+        $right = $length - 1;
+        // 比较
+        while ($left < $right) {
+            $sum = $nums[$i] + $nums[$left] + $nums[$right];
+           if ($sum < 0) {
+               $left++;
+           } elseif ($sum > 0) {
+               $right--;
+           } else {
+               array_push($result, [$nums[$i], $nums[$left], $nums[$right]]);
+               while ($left < $right && $nums[$left] == $nums[$left + 1]) $left++;
+               while ($left < $right && $nums[$right - 1] == $nums[$right]) $right--;
+               $left++;
+               $right--;
+            }
+        }
+    }
+
+    return $result;
+}
+```
+
+
+
 -----------------------
 * 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/0062.不同路径.md

@@ -279,9 +279,7 @@ Python：
 ```python
 class Solution: # 动态规划
     def uniquePaths(self, m: int, n: int) -> int:
-        dp = [[0 for i in range(n)] for j in range(m)]
-        for i in range(m): dp[i][0] = 1
-        for j in range(n): dp[0][j] = 1
+        dp = [[1 for i in range(n)] for j in range(m)]
         for i in range(1, m):
             for j in range(1, n):
                 dp[i][j] = dp[i][j - 1] + dp[i - 1][j]

problems/0070.爬楼梯.md

@@ -283,7 +283,18 @@ func climbStairs(n int) int {
     return dp[n]
 }
 ```
-
+Javascript:
+```Javascript
+var climbStairs = function(n) {
+    // dp[i] 为第 i 阶楼梯有多少种方法爬到楼顶
+    // dp[i] = dp[i - 1] + dp[i - 2]
+    let dp = [1 , 2]
+    for(let i = 2; i < n; i++) {
+        dp[i] = dp[i - 1] + dp[i - 2]
+    }
+    return dp[n - 1]
+};
+```
 
 
 -----------------------

problems/0098.验证二叉搜索树.md

@@ -376,6 +376,28 @@ func isBST(root *TreeNode, min, max int) bool {
 	return isBST(root.Left, min, root.Val) && isBST(root.Right, root.Val, max)
 }
 ```
+```go
+// 中序遍历解法
+func isValidBST(root *TreeNode) bool {
+    // 保存上一个指针
+    var prev *TreeNode
+    var travel func(node *TreeNode) bool
+    travel = func(node *TreeNode) bool {
+        if node == nil {
+            return true
+        }
+        leftRes := travel(node.Left)
+        // 当前值小于等于前一个节点的值，返回false
+        if prev != nil && node.Val <= prev.Val {
+            return false
+        }
+        prev = node
+        rightRes := travel(node.Right)
+        return leftRes && rightRes
+    }
+    return travel(root)
+}
+```
 
 JavaScript版本
 

problems/0226.翻转二叉树.md

@@ -205,6 +205,7 @@ public:
 Java：
 
 ```Java
+//DFS递归
 class Solution {
    /**
      * 前后序遍历都可以
@@ -226,6 +227,31 @@ class Solution {
         root.right = tmp;
     }
 }
+
+//BFS
+class Solution {
+    public TreeNode invertTree(TreeNode root) {
+        if (root == null) {return null;}
+        ArrayDeque<TreeNode> deque = new ArrayDeque<>();
+        deque.offer(root);
+        while (!deque.isEmpty()) {
+            int size = deque.size();
+            while (size-- > 0) {
+                TreeNode node = deque.poll();
+                swap(node);
+                if (node.left != null) {deque.offer(node.left);}
+                if (node.right != null) {deque.offer(node.right);}
+            }
+        }
+        return root;
+    }
+
+    public void swap(TreeNode root) {
+        TreeNode temp = root.left;
+        root.left = root.right;
+        root.right = temp;
+    }
+}
 ```
 
 Python：

problems/0235.二叉搜索树的最近公共祖先.md

@@ -313,65 +313,49 @@ func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
 }
 ```
 
-
-JavaScript版本
-> 递归
-
+JavaScript版本：
+1. 使用递归的方法
 ```javascript
-/**
- * Definition for a binary tree node.
- * function TreeNode(val) {
- *     this.val = val;
- *     this.left = this.right = null;
- * }
- */
-
-/**
- * @param {TreeNode} root
- * @param {TreeNode} p
- * @param {TreeNode} q
- * @return {TreeNode}
- */
 var lowestCommonAncestor = function(root, p, q) {
-    if(root.val > p.val && root.val > q.val)
-        return lowestCommonAncestor(root.left, p , q);
-    else if(root.val < p.val && root.val < q.val) 
-        return lowestCommonAncestor(root.right, p , q);
+    // 使用递归的方法
+    // 1. 使用给定的递归函数lowestCommonAncestor
+    // 2. 确定递归终止条件
+    if(root === null) {
+        return root;
+    }
+    if(root.val>p.val&&root.val>q.val) {
+        // 向左子树查询
+        let left = lowestCommonAncestor(root.left,p,q);
+        return left !== null&&left;
+    }
+    if(root.val<p.val&&root.val<q.val) {
+        // 向右子树查询
+        let right = lowestCommonAncestor(root.right,p,q);
+        return right !== null&&right;
+    }
     return root;
 };
 ```
-
-> 迭代
-
+2. 使用迭代的方法
 ```javascript
-/**
- * Definition for a binary tree node.
- * function TreeNode(val) {
- *     this.val = val;
- *     this.left = this.right = null;
- * }
- */
-
-/**
- * @param {TreeNode} root
- * @param {TreeNode} p
- * @param {TreeNode} q
- * @return {TreeNode}
- */
 var lowestCommonAncestor = function(root, p, q) {
-    while(1) {
-        if(root.val > p.val && root.val > q.val)
+    // 使用迭代的方法
+    while(root) {
+        if(root.val>p.val&&root.val>q.val) {
             root = root.left;
-        else if(root.val < p.val && root.val < q.val)
+        }else if(root.val<p.val&&root.val<q.val) {
             root = root.right;
-        else
-            break;
+        }else {
+            return root;
+        }
+        
     }
-    return root;
+    return null;
 };
 ```
 
 
+
 -----------------------
 * 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/0236.二叉树的最近公共祖先.md

@@ -311,35 +311,34 @@ func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
 }
 ```
 
-JavaScript版本
-
+JavaScript版本：
 ```javascript
-/**
- * Definition for a binary tree node.
- * function TreeNode(val) {
- *     this.val = val;
- *     this.left = this.right = null;
- * }
- */
-/**
- * @param {TreeNode} root
- * @param {TreeNode} p
- * @param {TreeNode} q
- * @return {TreeNode}
- */
 var lowestCommonAncestor = function(root, p, q) {
-    if(root === p || root === q || root === null)
-        return root;
-    let left = lowestCommonAncestor(root.left, p , q);
-    let right = lowestCommonAncestor(root.right, p, q);
-    if(left && right)
-        return root;
-    if(!left)
-        return right;
-    return left;
+    // 使用递归的方法
+    // 需要从下到上，所以使用后序遍历
+    // 1. 确定递归的函数
+    const travelTree = function(root,p,q) {
+        // 2. 确定递归终止条件
+        if(root === null || root === p||root === q) {
+            return root;
+        }
+        // 3. 确定递归单层逻辑
+        let left = travelTree(root.left,p,q);
+        let right = travelTree(root.right,p,q);
+        if(left !== null&&right !== null) {
+            return root;
+        }
+        if(left ===null) {
+            return right;
+        }
+        return left;
+    }
+   return  travelTree(root,p,q);
 };
 ```
 
+
+
 -----------------------
 * 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/0257.二叉树的所有路径.md

@@ -283,6 +283,7 @@ public:
 Java：
 
 ```Java
+//解法一
 class Solution {
     /**
      * 递归法
@@ -321,6 +322,52 @@ class Solution {
     }
 }
 
+//解法二（常规前序遍历，不用回溯），更容易理解
+class Solution {
+    public List<String> binaryTreePaths(TreeNode root) {
+        List<String> res = new ArrayList<>();
+        helper(root, new StringBuilder(), res);
+        return res;
+    }
+
+    public void helper(TreeNode root, StringBuilder sb, List<String> res) {
+        if (root == null) {return;}
+      	// 遇到叶子结点就放入当前路径到res集合中
+        if (root.left == null && root.right ==null) {
+            sb.append(root.val);
+            res.add(sb.toString());
+          	// 记得结束当前方法
+            return;
+        }
+        helper(root.left,new StringBuilder(sb).append(root.val + "->"),res);
+        helper(root.right,new StringBuilder(sb).append(root.val + "->"),res);
+    }
+}
+
+//针对解法二优化，思路本质是一样的
+class Solution {
+    public List<String> binaryTreePaths(TreeNode root) {
+        List<String> res = new ArrayList<>();
+        helper(root, "", res);
+        return res;
+    }
+
+    public void helper(TreeNode root, String path, List<String> res) {
+        if (root == null) {return;}
+      	// 由原始解法二可以知道，root的值肯定会下面某一个条件加入到path中，那么干脆直接在这一步加入即可
+        StringBuilder sb = new StringBuilder(path);
+        sb.append(root.val);
+        if (root.left == null && root.right ==null) {
+            res.add(sb.toString());
+        }else{
+          	// 如果是非叶子结点则还需要跟上一个 “->”
+            sb.append("->");
+            helper(root.left,sb.toString(),res);
+            helper(root.right,sb.toString(),res);
+        }
+    }
+}
+
 ```
 
 Python：
@@ -350,7 +397,7 @@ class Solution:
 
 ```
 Go：
-  
+
 ```go
 func binaryTreePaths(root *TreeNode) []string {
     res := make([]string, 0)