# 平衡二叉树

<p>给定一个二叉树，判断它是否是高度平衡的二叉树。</p>

<p>本题中，一棵高度平衡二叉树定义为：</p>

<blockquote>
<p>一个二叉树<em>每个节点 </em>的左右两个子树的高度差的绝对值不超过 1 。</p>
</blockquote>

<p> </p>

<p><strong>示例 1：</strong></p>
<img alt="" src="https://img-service.csdnimg.cn/img_convert/341009251c7788edc838975732064bf1.jpeg" style="width: 342px; height: 221px;" />
<pre>
<strong>输入：</strong>root = [3,9,20,null,null,15,7]
<strong>输出：</strong>true
</pre>

<p><strong>示例 2：</strong></p>
<img alt="" src="https://img-service.csdnimg.cn/img_convert/bce64ca4c0e80b345a5a286122112851.jpeg" style="width: 452px; height: 301px;" />
<pre>
<strong>输入：</strong>root = [1,2,2,3,3,null,null,4,4]
<strong>输出：</strong>false
</pre>

<p><strong>示例 3：</strong></p>

<pre>
<strong>输入：</strong>root = []
<strong>输出：</strong>true
</pre>

<p> </p>

<p><strong>提示：</strong></p>

<ul>
	<li>树中的节点数在范围 <code>[0, 5000]</code> 内</li>
	<li><code>-10<sup>4</sup> <= Node.val <= 10<sup>4</sup></code></li>
</ul>


## template

```cpp
#include <bits/stdc++.h>
using namespace std;

struct TreeNode
{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution
{
public:
    int depth(TreeNode *root)
    {
        if (root == NULL)
            return 0;
        int left = depth(root->left);
        int right = depth(root->right);
        return max(left, right) + 1;
    }

    bool isBalanced(TreeNode *root)
    {
        if (root == NULL)
            return true;
        if (abs(depth(root->left) - depth(root->right)) > 1)
            return false;
        else
            return isBalanced(root->left) && isBalanced(root->right);
    }
};
```

## 答案

```cpp

```

## 选项

### A

```cpp

```

### B

```cpp

```

### C

```cpp

```