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].

class Solution {
    int currSum, maxSum;
    int min, max;

    public boolean checkBST(TreeNode root) {
        if (root.left == null && root.right == null) {
            currSum = min = max = root.val;
            maxSum = Math.max(maxSum, root.val);
            return true;
        }
        int leftSum = 0, lowerLimit = Integer.MAX_VALUE;
        boolean isBST = true;
        if (root.left != null) {
            boolean leftAns = checkBST(root.left);
            if (!leftAns || max >= root.val)
                isBST = false;
            leftSum = currSum;
            lowerLimit = min;
        }
        int rightSum = 0, upperLimit = Integer.MIN_VALUE;
        if (root.right != null) {
            boolean rightAns = checkBST(root.right);
            if (!rightAns || min <= root.val)
                isBST = false;
            rightSum = currSum;
            upperLimit = max;
        }
        if (!isBST)
            return false;
        // Setting max sum
        currSum = leftSum + root.val + rightSum;
        maxSum = Math.max(maxSum, currSum);
        // Setting the upper and lower limits of this BST
        min = Math.min(lowerLimit, root.val);
        max = Math.max(upperLimit, root.val);
        return true;
    }

    public int maxSumBST(TreeNode root) {
        currSum = maxSum = 0;
        checkBST(root);
        return maxSum;
    }
}