[LeetCode] 333. Largest BST Subtree

Problem

Given a binary tree, find the largest subtree which is a Binary Search Tree (BST), where largest means subtree with largest number of nodes in it.

Note:
A subtree must include all of its descendants.

Example:

Input: [10,5,15,1,8,null,7]

   10 
   / \ 
  5  15 
 / \   \ 
1   8   7

Output: 3
Explanation: The Largest BST Subtree in this case is the highlighted one.
             The return value is the subtree's size, which is 3.

Follow up:
Can you figure out ways to solve it with O(n) time complexity?

Solution

First try:

class Solution {
    public int largestBSTSubtree(TreeNode root) {
        if (root == null) return 0;
        if (isBST(root)) return largestBSTSubtree(root.left)+largestBSTSubtree(root.right)+1;
        return Math.max(largestBSTSubtree(root.left), largestBSTSubtree(root.right));
    }
    private boolean isBST(TreeNode root) {
        if (root == null) return true;
        if (root.left != null && findMax(root.left) >= root.val) return false;
        if (root.right != null && findMin(root.right) <= root.val) return false;
        if (isBST(root.left) && isBST(root.right)) return true;
        return false;
    }
    private int findMax(TreeNode root) {
        while (root.right != null) root = root.right;
        return root.val;
    }
    private int findMin(TreeNode root) {
        while (root.left != null) root = root.left;
        return root.val;
    }
}

Optimized:

class Solution {
    public int largestBSTSubtree(TreeNode root) {
        if (root == null) return 0;
        if (isBST(root, null, null)) return largestBSTSubtree(root.left)+largestBSTSubtree(root.right)+1;
        return Math.max(largestBSTSubtree(root.left), largestBSTSubtree(root.right));
    }
    private boolean isBST(TreeNode root, Integer min, Integer max) {
        if (root == null) return true;
        if (min != null && min >= root.val) return false;
        if (max != null && max <= root.val) return false;
        if (isBST(root.left, min, root.val) && isBST(root.right, root.val, max)) return true;
        return false;
    }
}