# 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://assets.leetcode.com/uploads/2020/10/06/balance_1.jpg" 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://assets.leetcode.com/uploads/2020/10/06/balance_2.jpg" 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

```java
class TreeNode {
	int val;
	TreeNode left;
	TreeNode right;

	TreeNode(int x) {
		val = x;
	}
}

public class Solution {
	public boolean isBalanced(TreeNode root) {

		if (root == null) {
			return true;
		}

		return (Math.abs(maxDepth(root.left) - maxDepth(root.right)) <= 1) && isBalanced(root.left)
				&& isBalanced(root.right);
	}

	public int maxDepth(TreeNode root) {
		if (root == null) {
			return 0;
		}
		return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
	}
}

```

## 答案

```java

```

## 选项

### A

```java

```

### B

```java

```

### C

```java

```