# 124. Binary Tree Maximum Path Sum

### Leetcode

<https://leetcode.com/problems/binary-tree-maximum-path-sum/>

### 題目

A **path** in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence **at most once**. Note that the path does not need to pass through the root.

The **path sum** of a path is the sum of the node's values in the path.

Given the `root` of a binary tree, return *the maximum **path sum** of any **non-empty** path*.

&#x20;

**Example 1:**

![](https://assets.leetcode.com/uploads/2020/10/13/exx1.jpg)

```
Input: root = [1,2,3]
Output: 6
Explanation: The optimal path is 2 -> 1 -> 3 with a path sum of 2 + 1 + 3 = 6.
```

**Example 2:**

![](https://assets.leetcode.com/uploads/2020/10/13/exx2.jpg)

```
Input: root = [-10,9,20,null,null,15,7]
Output: 42
Explanation: The optimal path is 15 -> 20 -> 7 with a path sum of 15 + 20 + 7 = 42.
```

### 解答 <a href="#ti-jie" id="ti-jie"></a>

* **方法一**

```
var maxPathSum = function(root) {
    let maxSum = Number.MIN_SAFE_INTEGER;
        
    const getMaxSum = (node) => {
        if (!node) return 0;
        
        // 有可能子節點計算出來的最大值是負數，那就不要加上子節點的值計算，所以取 0
        const maxLeftValue = Math.max(getMaxSum(node.left), 0);
        const maxRightValue = Math.max(getMaxSum(node.right), 0);
        
        // 任一節點的 本身值+左右子節點值 可能為最大值
        const newPathSum = node.val + maxLeftValue + maxRightValue;
        maxSum = Math.max(newPathSum, maxSum)
        
        return node.val + Math.max(maxLeftValue, maxRightValue);
    }
    
    getMaxSum(root);
    return maxSum;
};
```

Runtime: 123 ms, faster than 39.37% of JavaScript online submissions for Binary Tree Maximum Path Sum.

Memory Usage: 51.6 MB, less than 53.36% of JavaScript online submissions for Binary Tree Maximum Path Sum.
