Problem

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

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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

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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

Solve

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# 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:
# 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:
        if not root :
            return 0
        res = root.val

        def dfs(node):
            nonlocal res
            if not node:
                return 0
            
            max_left = dfs(node.left) 
            max_right = dfs(node.right)
            if max_left + max_right + node.val > res:
                res = max_left + max_right + node.val

            
            return max( 0 , max_left + node.val , max_right + node.val )

        dfs(root)

        return res