# 二叉树中的最大路径和

<p><strong>路径</strong> 被定义为一条从树中任意节点出发，沿父节点-子节点连接，达到任意节点的序列。同一个节点在一条路径序列中 <strong>至多出现一次</strong> 。该路径<strong> 至少包含一个 </strong>节点，且不一定经过根节点。</p>

<p><strong>路径和</strong> 是路径中各节点值的总和。</p>

<p>给你一个二叉树的根节点 <code>root</code> ，返回其 <strong>最大路径和</strong> 。</p>

<p> </p>

<p><strong>示例 1：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/13/exx1.jpg" style="width: 322px; height: 182px;" />
<pre>
<strong>输入：</strong>root = [1,2,3]
<strong>输出：</strong>6
<strong>解释：</strong>最优路径是 2 -> 1 -> 3 ，路径和为 2 + 1 + 3 = 6</pre>

<p><strong>示例 2：</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/13/exx2.jpg" />
<pre>
<strong>输入：</strong>root = [-10,9,20,null,null,15,7]
<strong>输出：</strong>42
<strong>解释：</strong>最优路径是 15 -> 20 -> 7 ，路径和为 15 + 20 + 7 = 42
</pre>

<p> </p>

<p><strong>提示：</strong></p>

<ul>
	<li>树中节点数目范围是 <code>[1, 3 * 10<sup>4</sup>]</code></li>
	<li><code>-1000 <= Node.val <= 1000</code></li>
</ul>


## template

```java
class Solution {
	int maxSum = Integer.MIN_VALUE;

	public int maxPathSum(TreeNode root) {
		maxGain(root);
		return maxSum;
	}

	public int maxGain(TreeNode node) {
		if (node == null) {
			return 0;
		}

		int leftGain = Math.max(maxGain(node.left), 0);
		int rightGain = Math.max(maxGain(node.right), 0);

		int priceNewpath = node.val + leftGain + rightGain;

		maxSum = Math.max(maxSum, priceNewpath);

		return node.val + Math.max(leftGain, rightGain);
	}
}
```

## 答案

```java

```

## 选项

### A

```java

```

### B

```java

```

### C

```java

```