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   3Return 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;    }};