lintcode:Binary Tree Maximum Path Sum

Given a binary tree, find the maximum path sum.

The path may start and end at any node in the tree.

Example

Given the below binary tree,

       1      / \     2   3

Return 6.

/** * Definition of TreeNode: * class TreeNode { * public: *     int val; *     TreeNode *left, *right; *     TreeNode(int val) { *         this->val = val; *         this->left = this->right = NULL; *     } * } */class Solution {    private:    int max5(int a, int b, int c, int d, int e)    {        return max(max(max(a, b), max(c, d)), e);    }        int max3(int a, int b, int c)    {        return max(max(a, b), c);    }    int maxPathSumHelper(TreeNode *Node, int &maxSum) {                if (Node == NULL)            return 0;                    int leftMax  = maxPathSumHelper(Node->left, maxSum);        int rightMax = maxPathSumHelper(Node->right, maxSum);                maxSum = max5(maxSum, Node->val, Node->val+leftMax, Node->val+rightMax, Node->val+leftMax+rightMax);                return max3(Node->val, Node->val+leftMax, Node->val+rightMax);    }        public:    /**     * @param root: The root of binary tree.     * @return: An integer     */         int maxPathSum(TreeNode *root) {        // write your code here        int maxSum = -10000;        int leftMax  = maxPathSumHelper(root->left, maxSum);        int rightMax = maxPathSumHelper(root->right, maxSum);                return max5(maxSum, root->val, leftMax+root->val, rightMax+root->val, leftMax+rightMax+root->val);    }};