Merge pull request youngyangyang04#1075 from jobinben/master

添加 面试题02.07.链表相交.md JavaScript语言解法的注释

Author: youngyangyang04

problems/0110.平衡二叉树.md

@@ -604,7 +604,8 @@ func abs(a int)int{
 }
 ```
 
-## JavaScript 
+## JavaScript
+递归法： 
 ```javascript
 var isBalanced = function(root) {
     //还是用递归三部曲 + 后序遍历 左右中 当前左子树右子树高度相差大于1就返回-1
@@ -627,6 +628,48 @@ var isBalanced = function(root) {
 };
 ```
 
+迭代法：
+```javascript
+// 获取当前节点的高度
+var getHeight = function (curNode) {
+    let queue = [];
+    if (curNode !== null) queue.push(curNode); // 压入当前元素
+    let depth = 0, res = 0;
+    while (queue.length) {
+        let node = queue[queue.length - 1]; // 取出栈顶
+        if (node !== null) {
+            queue.pop();
+            queue.push(node);   // 中
+            queue.push(null);
+            depth++;
+            node.right && queue.push(node.right);   // 右
+            node.left && queue.push(node.left);     // 左
+        } else {
+            queue.pop();
+            node = queue[queue.length - 1];
+            queue.pop();
+            depth--;
+        }
+        res = res > depth ? res : depth;
+    }
+    return res;
+}
+var isBalanced = function (root) {
+    if (root === null) return true;
+    let queue = [root];
+    while (queue.length) {
+        let node = queue[queue.length - 1]; // 取出栈顶
+        queue.pop();    
+        if (Math.abs(getHeight(node.left) - getHeight(node.right)) > 1) {
+            return false;
+        }
+        node.right && queue.push(node.right);
+        node.left && queue.push(node.left);
+    }
+    return true;
+};
+```
+
 ## TypeScript
 
 ```typescript

problems/面试题02.07.链表相交.md

@@ -224,12 +224,14 @@ var getIntersectionNode = function(headA, headB) {
         lenA = getListLen(headA),
         lenB = getListLen(headB);
     if(lenA < lenB) {
+        // 下面交换变量注意加 “分号” ，两个数组交换变量在同一个作用域下时
+        // 如果不加分号，下面两条代码等同于一条代码: [curA, curB] = [lenB, lenA]
         [curA, curB] = [curB, curA];
         [lenA, lenB] = [lenB, lenA];
     }
     let i = lenA - lenB;
     while(i-- > 0) {
-        curA = curA.next
+        curA = curA.next;
     }
     while(curA && curA !== curB) {
         curA = curA.next;