标签：path leetcode com 最大路 fine nod connect 左右 序列
Given a non-empty binary tree, find the maximum path sum.
For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path must contain at least one node and does not need to go through the root.
Example 1:
Input: [1,2,3]
1
/ 2 3
Output: 6
Example 2:
Input: [-10,9,20,null,null,15,7] -10 / 9 20 / 15 7 Output: 42
给定一颗二叉树，求其最大路径，路径就是从任意节点出发，到达任意节点的序列，注意路径至少要有一个节点。
这道题和下面两道题做法相同，都是求二叉树的路径问题，可以先回顾下面两个问题。
LeetCode 543. Diameter of Binary Tree 二叉树的直径 (C++/Java)
LeetCode 687. Longest Univalue Path 最长同值路径 (C++/Java)
那么这道题还是从根节点递归求解，当前结点的最大路径，是当前节点的值加上左右孩子的最大路径，同时和全局的最大值进行比较，更新最大值，而作为返回值时，需要在左右孩子中选取最大值加上当前结点的值同时和0比较，选取最大值，因为节点存在负数，那么路径为负数显然是不对的，所以对于值为负数的子树我们向上层返回0即可。
C++
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int maxPathSum(TreeNode* root) { if(root == nullptr) return 0; int res = INT_MIN; maxPathSum(root, res); return res; } private: int maxPathSum(TreeNode* root, int& res){ if(root == nullptr) return 0; int l = maxPathSum(root->left, res); int r = maxPathSum(root->right, res); int sum = l + r + root->val; res = max(sum, res); return max(max(l, r) + root->val, 0); } };
Java
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public int maxPathSum(TreeNode root) { if(root == null) return 0; res = Integer.MIN_VALUE; maxPath(root); return res; } private int res; private int maxPath(TreeNode root) { if(root == null) return 0; int l = maxPath(root.left); int r = maxPath(root.right); int sum = l + r + root.val; res = Math.max(sum, res); return Math.max(Math.max(l, r) + root.val, 0); } }
LeetCode 124. Binary Tree Maximum Path Sum 二叉树中的最大路径和 (C++/Java)
标签：path leetcode com 最大路 fine nod connect 左右 序列
原文地址：https://www.cnblogs.com/silentteller/p/12376661.html
