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.

Input and output examples

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

Solution

# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution(object): def maxPathSum(self, root): """ :type root: TreeNode :rtype: int """ self.max_sum = float('-inf') self.max_gain(root) return self.max_sum def max_gain(self, node): if not node: return 0 left_gain = max(self.max_gain(node.left), 0) right_gain = max(self.max_gain(node.right), 0) price_newpath = node.val + left_gain + right_gain self.max_sum = max(self.max_sum, price_newpath) return node.val + max(left_gain, right_gain)

Test the the solution:

class TreeNode(object): def __init__(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right

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