[Leetcode] 0124. Binary Tree Maximum Path Sum

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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:

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 * 10 ⁴ ].
-1000 <= Node.val <= 1000

Table of contents

Python

				
					# time complexity: O(n) # space complexity: O(n) from typing import Optional 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: maxPath = float('-inf') def dfs(node: Optional[TreeNode]): nonlocal maxPath if node is None: return 0 pathLeft = max(dfs(node.left), 0) pathRight = max(dfs(node.right), 0) maxPath = max(maxPath, pathLeft + pathRight + node.val) return max(pathLeft + node.val, pathRight + node.val) dfs(root) return maxPath root = TreeNode(1) root.left = TreeNode(2) root.right = TreeNode(3) print(Solution().maxPathSum(root))

Python

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