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，計算每個節點的左支路值與右支路值和本節點值，比較最大值並更新加上該節點的支路
最大值(並與0比較，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;
    }
};