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.

/**
* 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) {
maxSum = INT_MIN;
pathSum(root);
return maxSum;
}
int pathSum(TreeNode* root){
if(!root) return 0;

int leftSum = pathSum(root->left);
int rightSum = pathSum(root->right);
int sum = root->val;
if(leftSum > 0) sum += leftSum;
if(rightSum > 0) sum += rightSum;
maxSum = max(sum, maxSum);
return max(leftSum, rightSum) > 0 ? max(leftSum, rightSum) + root->val : root->val;
}
private:
int maxSum;
};