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124.二叉树中的最大路径和 ​

dfs, https://leetcode.cn/problems/binary-tree-maximum-path-sum/

二叉树中的 路径 被定义为一条节点序列，序列中每对相邻节点之间都存在一条边。同一个节点在一条路径序列中 至多出现一次 。该路径 至少包含一个 节点，且不一定经过根节点。

路径和 是路径中各节点值的总和。

给你一个二叉树的根节点 root ，返回其 最大路径和 。

示例 1：

img
输入：root = [1,2,3]
输出：6
解释：最优路径是 2 -> 1 -> 3 ，路径和为 2 + 1 + 3 = 6

示例 2：

img
输入：root = [-10,9,20,null,null,15,7]
输出：42
解释：最优路径是 15 -> 20 -> 7 ，路径和为 15 + 20 + 7 = 42

提示：

  • 树中节点数目范围是 [1, 3 * 10^4]
  • -1000 <= Node.val <= 1000
python
# 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:
        self.max_sum = float('-inf')

        def max_gain(node):
            if not node:
                return 0

            # Recursively get the maximum gain from the left and right subtrees
            left_gain = max(max_gain(node.left), 0)
            right_gain = max(max_gain(node.right), 0)

            # Calculate the price of the current path
            current_path_sum = node.val + left_gain + right_gain

            # Update the global maximum path sum
            self.max_sum = max(self.max_sum, current_path_sum)

            # Return the maximum gain the current node can contribute to its parent
            return node.val + max(left_gain, right_gain)

        max_gain(root)
        return self.max_sum