Merge branch 'youngyangyang04:master' into master

problems/0051.N皇后.md

@@ -341,7 +341,7 @@ class Solution {
 
     public boolean isValid(int row, int col, int n, char[][] chessboard) {
         // 检查列
-        for (int i=0; i<n; ++i) {
+        for (int i=0; i<row; ++i) { // 相当于剪枝
             if (chessboard[i][col] == 'Q') {
                 return false;
             }

problems/0104.二叉树的最大深度.md

@@ -505,21 +505,51 @@ var maxdepth = function(root) {
 
 二叉树最大深度层级遍历
 ```javascript
-var maxdepth = function(root) {
-    //使用递归的方法 递归三部曲
-    //1. 确定递归函数的参数和返回值
-    const getdepth=function(node){
-    //2. 确定终止条件
-        if(node===null){
-            return 0;
+var maxDepth = function(root) {
+    if(!root) return 0
+    let count = 0
+    const queue = [root]
+    while(queue.length) {
+        let size = queue.length
+        /* 层数+1 */
+        count++
+        while(size--) {
+            let node = queue.shift();
+            node.left && queue.push(node.left);
+            node.right && queue.push(node.right);
         }
-    //3. 确定单层逻辑
-        let leftdepth=getdepth(node.left);
-        let rightdepth=getdepth(node.right);
-        let depth=1+math.max(leftdepth,rightdepth);
-        return depth;
     }
-    return getDepth(root);
+    return count
+};
+```
+
+N叉树的最大深度  递归写法
+```js
+var maxDepth = function(root) {
+    if(!root) return 0
+    let depth = 0
+    for(let node of root.children) {
+        depth = Math.max(depth, maxDepth(node))
+    }
+    return depth + 1
+}
+```
+
+N叉树的最大深度  层序遍历
+```js
+var maxDepth = function(root) {
+    if(!root) return 0
+    let count = 0
+    let queue = [root]
+    while(queue.length) {
+        let size = queue.length
+        count++
+        while(size--) {
+            let node = queue.shift()
+            node && (queue = [...queue, ...node.children])
+        }
+    }
+    return count
 };
 ```
 

problems/1002.查找常用字符.md

@@ -181,7 +181,7 @@ class Solution:
         # 统计除第一个字符串外字符的出现频率
         for i in range(1, len(words)):
             hashOtherStr = [0] * 26
-            for j in range(len(words[0])):
+            for j in range(len(words[i])):
                 hashOtherStr[ord(words[i][j]) - ord('a')] += 1
             # 更新hash，保证hash里统计26个字符在所有字符串里出现的最小次数
             for k in range(26):