LeetCode: 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.

For example:
Given the below binary tree,

       1      / \     2   3

Return 6.

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    int maxPathSum(TreeNode *root) {        max = INT_MIN;        buildMaxMap(root);        return max;                    }    int buildMaxMap(TreeNode* node)    {        if(node == NULL)            return 0;        int maxL = buildMaxMap(node->left);        int maxR = buildMaxMap(node->right);        int maxNode = std::max(std::max(maxL, maxR) + node->val, node->val);        int curMax = std::max(std::max(maxL + maxR + node->val, node->val), maxNode);        if(curMax > max)            max = curMax;        return maxNode;            }private:int max;    };

Round 2:

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    int maxPathSum(TreeNode *root) {        result = INT_MIN;        helper(root);        return result;    }private:    int helper(TreeNode *root)    {        if(root == NULL)            return 0;        int left = helper(root->left);        int right = helper(root->right);        int curMax = 0, localMax = 0;        curMax = std::max(std::max(left, right) + root->val, root->val);        localMax = std::max(curMax, left + right + root->val);        if(localMax > result)            result = localMax;        return curMax;            }    int result;    };