# 124. Binary Tree Maximum Path Sum

#### Hard

***

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*.

 

**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.
```

 

**Constraints:**

* The number of nodes in the tree is in the range `[1, 3 * 104]`.
* `-1000 <= Node.val <= 1000`

```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:
        self.result = root.val
        self.traverse(root)
        return self.result
    
    def traverse(self, root):
        if not root:
            return 0
        left = self.traverse(root.left)
        right = self.traverse(root.right)
        self.result = max(self.result, max(left,0) + max(right,0) + root.val)
        return max(max(left,0), max(right,0)) + root.val
        
```