二叉树中的最大路径和
Binary Tree Maximum Path Sum
题目详情
问题：二叉树中的最大路径和
考察：树、递归
来源：Citadel
链接：https://www.jointaro.com/interviews/questions/binary-tree-maximum-path-sum/
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 * 104]. -1000 <= Node.val <= 1000
解析
思路：后序遍历每个节点，计算从该节点向父节点延伸的最大单边贡献；负贡献直接舍弃为 0。用左贡献 + 当前值 + 右贡献更新全局最大路径和。
复杂度：每个节点访问一次，时间 O(n)，递归栈 O(h)。