124_binaryTreeMaximumPathSum

/*
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 must contain at least one node and does not need to go through the root.

For example:
Given the below binary tree,

1
/ \
2   3
Return 6.
自下而上，计算每个节点的左支路值与右支路值和本节点值，比较最大值并更新加上该节点的支路
最大值(并与0比较，取大的)
*/

/**
* 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) {
int sum=INT_MIN;
int tmp=findMax(root,sum);
sum=max(tmp,sum);
return sum;
}
int findMax(TreeNode* root,int &sum){
if(!root) return 0;
int leftSum=max(0,findMax(root->left,sum));
int rightSum=max(0,findMax(root->right,sum));
sum = (leftSum+rightSum+root->val) > sum ? leftSum+rightSum+root->val : sum;
return leftSum > rightSum ? leftSum+root->val : rightSum+root->val;
}
};