Binary Tree maximum path sum

Given a non-empty 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.

Example 1:

Input: [1,2,3]

       1
      / \
     2   3

Output: 6

Example 2:

Input: [-10,9,20,null,null,15,7]

   -10
   / \
  9  20
    /  \
   15   7

Output: 42

/**
 * 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) {
        int ans = INT_MIN;
        int sumWithRoot = helper(root, ans);
        return ans != INT_MIN ? ans : 0;
    }
   
    int helper(TreeNode *root, int& ans) {
        if (root == NULL) {
            return 0;
        }

        int leftSum = helper(root -> left, ans);
        int rightSum = helper(root -> right, ans);
       
        int sumWithRoot =  root -> val;
        if (leftSum > 0) {
            sumWithRoot += leftSum;
        }
        if (rightSum > 0) {
            sumWithRoot += rightSum;
        }
        ans = max(ans, sumWithRoot);
        if (max(leftSum, rightSum) > 0) {
            return max(leftSum, rightSum) + root -> val;
        } else {
            return root -> val;
        }
    }
};

===========

class Solution {

public:

    int maxPathSum(TreeNode* root) {

        int ans = INT_MIN;

        helper(root, ans);

        return ans;

    }

    

    int helper(TreeNode* root, int& ans) {

        if (root == NULL) {

            return 0;

        }

        int left_sum = helper(root -> left, ans);

        int right_sum = helper(root -> right, ans);

        ans = max(ans, left_sum + right_sum + root -> val);

        int sum = (left_sum > right_sum ? left_sum : right_sum) + root -> val;

        return sum > 0 ? sum : 0;

    }

};