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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.
/**
* 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 maxroute(TreeNode* root)
{
if(root==NULL)
return 0;
int left=maxroute(root->left);
int right=maxroute(root->right);
int tmp=left>right?left:right;
res=(root->val)>res?root->val:res;
if(tmp>left+right&&root->val+tmp>res)
res=root->val+tmp;
else if(root->val+left+right>res)
res=root->val+left+right;
return (tmp>0?root->val+tmp:root->val);
}
int maxPathSum(TreeNode* root) {
if (root==NULL)
return 0;
if(root->left==NULL&&root->right==NULL)
return root->val;
res=root->val;
int cc= maxroute( root);
return res;
}
int res;
};标签:
原文地址:http://blog.csdn.net/zhouyelihua/article/details/45629025