LeetCode [1373] Maximum Sum BST in Binary Tree

1373. Maximum Sum BST in Binary Tree

Hard

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].

/**
 * 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 {
    int ret = 0;
    public int maxSumBST(TreeNode root) {
        int[] v = valid(root);
        return ret;
    }
    
    //{isBST 0/1, small, largest, sum}
    int[] valid(TreeNode root){
        if(root == null){
            return new int[]{1, Integer.MAX_VALUE, Integer.MIN_VALUE, 0};
        }
        int[] left = valid(root.left);
        int[] right = valid(root.right);
        boolean isBST = left[0]==1 &&
                        right[0]==1 &&
                        left[2] < root.val &&
                        root.val < right[1];
        int sum = left[3]+right[3]+root.val;
        if(isBST){
            ret = Math.max(ret, sum);
        }
        return new int[]{isBST?1:0, Math.min(root.val,left[1]), Math.max(root.val, right[2]), isBST?sum:0};
    }
}