---
comments: true
difficulty: 困难
edit_url: https://github.com/doocs/leetcode/edit/main/solution/0100-0199/0124.Binary%20Tree%20Maximum%20Path%20Sum/README.md
tags:
    - 树
    - 深度优先搜索
    - 动态规划
    - 二叉树
---

<!-- problem:start -->

# [124. 二叉树中的最大路径和](https://leetcode.cn/problems/binary-tree-maximum-path-sum)

[English Version](/solution/0100-0199/0124.Binary%20Tree%20Maximum%20Path%20Sum/README_EN.md)

## 题目描述

<!-- description:start -->

<p>二叉树中的<strong> 路径</strong> 被定义为一条节点序列，序列中每对相邻节点之间都存在一条边。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>&nbsp;</p>

<p><strong>示例 1：</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0124.Binary%20Tree%20Maximum%20Path%20Sum/images/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 -&gt; 1 -&gt; 3 ，路径和为 2 + 1 + 3 = 6</pre>

<p><strong>示例 2：</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0124.Binary%20Tree%20Maximum%20Path%20Sum/images/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 -&gt; 20 -&gt; 7 ，路径和为 15 + 20 + 7 = 42
</pre>

<p>&nbsp;</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 &lt;= Node.val &lt;= 1000</code></li>
</ul>

<!-- description:end -->

## 解法

<!-- solution:start -->

### 方法一：递归

我们思考二叉树递归问题的经典套路：

1. 终止条件（何时终止递归）
2. 递归处理左右子树
3. 合并左右子树的计算结果

对于本题，我们设计一个函数 $dfs(root)$，它返回以 $root$ 为根节点的二叉树的最大路径和。

函数 $dfs(root)$ 的执行逻辑如下：

如果 $root$ 不存在，那么 $dfs(root)$ 返回 $0$；

否则，我们递归计算 $root$ 的左子树和右子树的最大路径和，分别记为 $left$ 和 $right$。如果 $left$ 小于 $0$，那么我们将其置为 $0$，同理，如果 $right$ 小于 $0$，那么我们将其置为 $0$。

然后，我们用 $root.val + left + right$ 更新答案。最后，函数返回 $root.val + \max(left, right)$。

在主函数中，我们调用 $dfs(root)$，即可得到每个节点的最大路径和，其中的最大值即为答案。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点数。

<!-- tabs:start -->

#### Python3

```python
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def maxPathSum(self, root: Optional[TreeNode]) -> int:
        def dfs(root: Optional[TreeNode]) -> int:
            if root is None:
                return 0
            left = max(0, dfs(root.left))
            right = max(0, dfs(root.right))
            nonlocal ans
            ans = max(ans, root.val + left + right)
            return root.val + max(left, right)

        ans = -inf
        dfs(root)
        return ans
```

#### Java

```java
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private int ans = -1001;

    public int maxPathSum(TreeNode root) {
        dfs(root);
        return ans;
    }

    private int dfs(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int left = Math.max(0, dfs(root.left));
        int right = Math.max(0, dfs(root.right));
        ans = Math.max(ans, root.val + left + right);
        return root.val + Math.max(left, right);
    }
}
```

#### C++

```cpp
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int maxPathSum(TreeNode* root) {
        int ans = -1001;
        function<int(TreeNode*)> dfs = [&](TreeNode* root) {
            if (!root) {
                return 0;
            }
            int left = max(0, dfs(root->left));
            int right = max(0, dfs(root->right));
            ans = max(ans, left + right + root->val);
            return root->val + max(left, right);
        };
        dfs(root);
        return ans;
    }
};
```

#### Go

```go
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func maxPathSum(root *TreeNode) int {
	ans := -1001
	var dfs func(*TreeNode) int
	dfs = func(root *TreeNode) int {
		if root == nil {
			return 0
		}
		left := max(0, dfs(root.Left))
		right := max(0, dfs(root.Right))
		ans = max(ans, left+right+root.Val)
		return max(left, right) + root.Val
	}
	dfs(root)
	return ans
}
```

#### TypeScript

```ts
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function maxPathSum(root: TreeNode | null): number {
    let ans = -1001;
    const dfs = (root: TreeNode | null): number => {
        if (!root) {
            return 0;
        }
        const left = Math.max(0, dfs(root.left));
        const right = Math.max(0, dfs(root.right));
        ans = Math.max(ans, left + right + root.val);
        return Math.max(left, right) + root.val;
    };
    dfs(root);
    return ans;
}
```

#### Rust

```rust
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
    fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, res: &mut i32) -> i32 {
        if root.is_none() {
            return 0;
        }
        let node = root.as_ref().unwrap().borrow();
        let left = (0).max(Self::dfs(&node.left, res));
        let right = (0).max(Self::dfs(&node.right, res));
        *res = (node.val + left + right).max(*res);
        node.val + left.max(right)
    }

    pub fn max_path_sum(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        let mut res = -1000;
        Self::dfs(&root, &mut res);
        res
    }
}
```

#### JavaScript

```js
/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var maxPathSum = function (root) {
    let ans = -1001;
    const dfs = root => {
        if (!root) {
            return 0;
        }
        const left = Math.max(0, dfs(root.left));
        const right = Math.max(0, dfs(root.right));
        ans = Math.max(ans, left + right + root.val);
        return Math.max(left, right) + root.val;
    };
    dfs(root);
    return ans;
};
```

#### C#

```cs
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    private int ans = -1001;

    public int MaxPathSum(TreeNode root) {
        dfs(root);
        return ans;
    }

    private int dfs(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int left = Math.Max(0, dfs(root.left));
        int right = Math.Max(0, dfs(root.right));
        ans = Math.Max(ans, left + right + root.val);
        return root.val + Math.Max(left, right);
    }
}
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- problem:end -->