# 124 Binary Tree Maximum Path Sum

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:

``````   1
/ \
2   3

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:

`````` -10
/ \
9   20
/  \
15   7

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`
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 `````` ``````# 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 __init__(self): self.max_sum = float('-inf') def maxPathSum(self, root: Optional[TreeNode]) -> int: self.traverse(root) return self.max_sum def traverse(self, root: Optional[TreeNode]) -> int: if not root: return 0 left = self.traverse(root.left) right = self.traverse(root.right) self.max_sum = max(self.max_sum, left + right + root.val) return max(max(left, right) + root.val, 0)``````