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.

Runtime: 32ms.

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     int maxPathSum(TreeNode* root) {
13         maxSum = INT_MIN;
14         pathSum(root);
15         return maxSum;
16     }
17     int pathSum(TreeNode* root){
18         if(!root) return 0;
19         
20         int leftSum = pathSum(root->left);
21         int rightSum = pathSum(root->right);
22         int sum = root->val;
23         if(leftSum > 0) sum += leftSum;
24         if(rightSum > 0) sum += rightSum;
25         maxSum = max(sum, maxSum);
26         return max(leftSum, rightSum) > 0 ? max(leftSum, rightSum) + root->val : root->val;
27     }
28 private:
29     int maxSum;
30 };