# 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

```python
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None


class Solution(object):
    def isBalanced(self, root):
        if not root:
            return True
        left_depth = self.get_depth(root.left)
        right_depth = self.get_depth(root.right)
        if abs(left_depth - right_depth) > 1:
            return False

        else:
            return self.isBalanced(root.left) and self.isBalanced(root.right)

    def get_depth(self, root):

        if root is None:
            return 0
        else:
            return max(self.get_depth(root.left), self.get_depth(root.right)) + 1

```

## 答案

```python

```

## 选项

### A

```python

```

### B

```python

```

### C

```python

```