1. 题目信息

给定一个非空二叉树，返回其最大路径和。

本题中，路径被定义为一条从树中任意节点出发，达到任意节点的序列。该路径至少包含一个节点，且不一定经过根节点。

示例 1:输入: [1,2,3]1/ \2   3输出: 6
示例 2:输入: [-10,9,20,null,null,15,7]-10/ \9  20/  \15   7输出: 42

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/binary-tree-maximum-path-sum
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2. 解题

• 以当前节点为根的子树的最大值

  int curmax = root->val;
  if(left > 0)curmax += left;
  if(right > 0)curmax += right;
  

• 但是返回给上层使用的时候只能保留一边子树

  max(root->val, max(left+root->val, right+root->val))
  

/*** 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 {	//C++
public:int maxPathSum(TreeNode* root) {int maxanswer = INT_MIN;maxsum(root, maxanswer);return maxanswer;}int maxsum(TreeNode* root, int &maxanswer) {if(root == NULL)return 0;int left = maxsum(root->left,maxanswer);int right = maxsum(root->right,maxanswer);int curmax = root->val;if(left > 0)curmax += left;if(right > 0)curmax += right;if(curmax > maxanswer)maxanswer = curmax;return max(root->val,max(left+root->val,right+root->val));}
};

• python3 解答

class Solution:# py3def maxPathSum(self, root: TreeNode) -> int:self.maxsum = float('-inf')def dfs(root):if not root:return 0l = dfs(root.left)r = dfs(root.right)v = root.valcurmax = vif l>0:curmax += lif r>0:curmax += rself.maxsum = max(curmax, self.maxsum)return max(v, max(v+l, v+r))dfs(root)return self.maxsum