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Given a binary tree, find the maximum path sum.
The path may start and end at any node in the tree.
For example:
Given the below binary tree,
1 / 2 3
Return 6
.
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 int maxSum=INT_MIN; 13 int maxPathSum(TreeNode *root) { 14 15 DFS(root); 16 return maxSum; 17 } 18 19 int DFS(TreeNode *root) 20 { 21 if(root==NULL) 22 { 23 return 0; 24 } 25 26 int left=DFS(root->left); 27 int right=DFS(root->right); 28 29 int sum=root->val; 30 if(left>0) sum+=left; 31 if(right>0) sum+=right; 32 if(maxSum<sum) maxSum=sum; 33 34 return (left>0||right>0)?root->val+max(left,right):root->val; 35 } 36 };
【leetcode】Binary Tree Maximum Path Sum
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原文地址:http://www.cnblogs.com/reachteam/p/4199301.html