Maximum Sum BST in Binary Tree

Given a binary tree root, the task is to return the maximum sum of all keys of any sub-tree which is also a Binary Search Tree (BST).

Assume a BST is defined as follows:

The left subtree of a node contains only nodes with keys less than the node's key.
The right subtree of a node contains only nodes with keys greater than the node's key.
Both the left and right subtrees must also be binary search trees.

Example 1:

Input: root = [1,4,3,2,4,2,5,null,null,null,null,null,null,4,6]
Output: 20
Explanation: Maximum sum in a valid Binary search tree is obtained in root node with key equal to 3.

Example 2:

Input: root = [4,3,null,1,2]
Output: 2
Explanation: Maximum sum in a valid Binary search tree is obtained in a single root node with key equal to 2.

Example 3:

Input: root = [-4,-2,-5]
Output: 0
Explanation: All values are negatives. Return an empty BST.

Example 4:

Input: root = [2,1,3]
Output: 6

Example 5:

Input: root = [5,4,8,3,null,6,3]
Output: 7

Constraints:

The given binary tree will have between 1 and 40000 nodes.
Each node's value is between [-4 * 10^4 , 4 * 10^4].

Solution:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int maxSumBST(TreeNode root) {
        int[] maxSum = new int[]{0};
        helper(root, maxSum);
        return maxSum[0];
    }
    
    private int[] helper(TreeNode root, int[] maxSum) {
        // [isBST, min, max, sum]
        if (root == null) {
            return new int[]{1, Integer.MAX_VALUE, Integer.MIN_VALUE, 0};
        }
        int[] left = helper(root.left, maxSum);
        int[] right = helper(root.right, maxSum);
        int[] ret = new int[4];
        if (left[0] == 1 && right[0] == 1 && root.val > left[2] && root.val < right[1]) {
            ret[0] = 1;
            ret[1] = Math.min(left[1], root.val);
            ret[2] = Math.max(right[2], root.val);
            ret[3] = root.val + left[3] + right[3];
            maxSum[0] = Math.max(maxSum[0], ret[3]);
        }
        return ret;
    }
}