标签：
Given a 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 does not need to go through the root.
For example:
Given the below binary tree,
1
/ 2 3
Return 6.
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int findMax(TreeNode *root,int &res)
{
if(root==NULL)return 0;
int value=root->val;
int left=findMax(root->left,res);
int right=findMax(root->right,res);
value+=left>0?left:0;
value+=right>0?right:0;
res=value>res?value:res;
return max(root->val,max(root->val+left,root->val+right));
}
int maxPathSum(TreeNode *root) {
int res=INT_MIN;
findMax(root,res);
return res;
}
};
二叉树最大路径和-Binary Tree Maximum Path Sum
标签：
原文地址：http://www.cnblogs.com/Vae1990Silence/p/4830620.html
