From 18ba70884914ed10944da16d60bba1fda610262f Mon Sep 17 00:00:00 2001
From: Arthur <gpan20020205@gmail.com>
Date: Thu, 4 Nov 2021 10:56:41 +0000
Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200110.=E5=B9=B3=E8=A1=A1?=
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---
 problems/0110.平衡二叉树.md | 63 ++++++++++++++++++++++++++++++++
 1 file changed, 63 insertions(+)

diff --git a/problems/0110.平衡二叉树.md b/problems/0110.平衡二叉树.md
index 2e8c239f..f3d70a5c 100644
--- a/problems/0110.平衡二叉树.md
+++ b/problems/0110.平衡二叉树.md
@@ -666,6 +666,69 @@ bool isBalanced(struct TreeNode* root) {
 }
 ```
 
+迭代法：
+```c
+//计算结点深度
+int getDepth(struct TreeNode* node) {
+    struct TreeNode** stack = (struct TreeNode**)malloc(sizeof(struct TreeNode*) * 10000);
+    int stackTop = 0;
+    if(node)
+        stack[stackTop++] = node;
+    int result = 0;
+    int depth = 0;
+    
+    while(stackTop) {
+        struct TreeNode* tempNode = stack[--stackTop];
+        if(tempNode) {
+            depth++;
+            stack[stackTop++] = tempNode;
+            stack[stackTop++] = NULL;
+            if(tempNode->left)
+                stack[stackTop++] = tempNode->left;
+            if(tempNode->right)
+                stack[stackTop++] = tempNode->right;
+            result = result > depth ? result : depth;
+        }
+        else {
+            tempNode = stack[--stackTop];
+            depth--;
+        }
+    }    
+
+    return result;
+}
+
+bool isBalanced(struct TreeNode* root){
+    //开辟栈空间
+    struct TreeNode** stack = (struct TreeNode**)malloc(sizeof(struct TreeNode*) * 10000);
+    int stackTop = 0;
+    
+    //若根节点不存在，返回True
+    if(!root)
+        return 1;
+    
+    //将根节点入栈
+    stack[stackTop++] = root;
+    //当栈中有元素时，进行遍历
+    while(stackTop) {
+        //将栈顶元素出栈
+        struct TreeNode* node = stack[--stackTop];
+        //计算左右子树的深度
+        int diff = getDepth(node->right) - getDepth(node->left);
+        //若深度的绝对值大于1，返回False
+        if(diff > 1 || diff < -1) 
+            return 0;
+        //如果栈顶结点有左右结点，将左右结点入栈
+        if(node->left)
+            stack[stackTop++] = node->left;
+        if(node->right)
+            stack[stackTop++] = node->right;
+    }
+    //若二叉树遍历结束后没有返回False，则返回True
+    return 1;
+}
+```
+
 -----------------------
 * 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)