# LeetCode 124. Binary Tree Maximum Path Sum

## Description

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

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 path.

Example 1:

```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:

```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`

## Explanation

Use a helper function to find the max path value from a node. Traverse each node to globally find the maximum path sum of the whole tree.

## Python Solution

``````# 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: TreeNode) -> int:

self.max_sum = -sys.maxsize
self.helper(root)
return self.max_sum

def helper(self, root):
if root == None:
return 0

left = max(self.helper(root.left), 0)
right = max(self.helper(root.right), 0)

self.max_sum = max(self.max_sum, left + right + root.val)

return max(left, right) + root.val
``````
• Time Complexity: O(N).
• Space Complexity: O(N).