leetcode 124. Binary Tree Maximum Path Sum

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.

/** * 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)     {        maxPath = INT_MIN;        helper(root);        return maxPath;    }private:    int maxPath;    int helper(TreeNode* root) //返回 包括自己在内的最大值 给上一层节点    {        if (!root->left && !root->right) //左右节点都没有        {            maxPath = max(maxPath, root->val);            return root->val;        }        else if (!root->left && root->right)        {            int rc = helper(root->right);            rc = rc > 0 ? rc : 0;            maxPath = max(maxPath, root->val + rc);            return root->val + rc;        }        else if (root->left && !root->right)        {            int rl = helper(root->left);            rl = rl > 0 ? rl : 0;            maxPath = max(maxPath, root->val + rl);            return root->val + rl;        }        else        {            int rc = helper(root->right);            int lc = helper(root->left);            rc = rc > 0 ? rc : 0;            lc = lc > 0 ? lc : 0;            maxPath = max(maxPath, root->val + rc + lc); //弯曲的情况，当前节点是弯曲点            return root->val + max(rc, lc);        }    }    };