333 Largest BST subtree

https://leetcode.com/problems/largest-bst-subtree/

对每个node进行bst validate（注意这里validate的方法）， 如果遇到bst再get一下num就行了

/**
 * 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 {
    private int ans = 0;
    public int largestBSTSubtree(TreeNode root) {
        largest(root);
        return ans;
    }
    private void largest(TreeNode root) {
        if (root == null) return;
        if (validate(root, Integer.MIN_VALUE, Integer.MAX_VALUE)) {
            ans = Math.max(getNum(root), ans);
        } else {
            largest(root.left);
            largest(root.right);
        }
    }
    private boolean validate(TreeNode root, int min, int max) {
        if (root == null) return true;
        if (root.val <= min || root.val >= max) return false;
        return validate(root.left, min, root.val) && validate(root.right, root.val, max);
    }
    private int getNum(TreeNode root) {
        if (root == null) return 0;
        return getNum(root.left) + getNum(root.right) + 1;
    }
}

这样时间复杂度是O(n^2)，也可以记忆化搜索，记录每个node对应的tree的最大值最小值和size，这样就不用重复遍历某些node了

// Each node will return min node value, max node value, max size
class NodeValue {
    public int maxNode, minNode, maxSize;
    
    NodeValue(int minNode, int maxNode, int maxSize) {
        this.maxNode = maxNode;
        this.minNode = minNode;
        this.maxSize = maxSize;
    }
};

class Solution {
    public NodeValue largestBSTSubtreeHelper(TreeNode root) {
        // An empty tree is a BST of size 0.
        if (root == null) {
            return new NodeValue(Integer.MAX_VALUE, Integer.MIN_VALUE, 0);
        }
        
        // Get values from left and right subtree of current tree.
        NodeValue left = largestBSTSubtreeHelper(root.left);
        NodeValue right = largestBSTSubtreeHelper(root.right);
        
        // Current node is greater than max in left AND smaller than min in right, it is a BST.
        if (left.maxNode < root.val && root.val < right.minNode) {
            // It is a BST.
            return new NodeValue(Math.min(root.val, left.minNode), Math.max(root.val, right.maxNode), 
                                left.maxSize + right.maxSize + 1);
        }
        
        // Otherwise, return [-inf, inf] so that parent can't be valid BST
        return new NodeValue(Integer.MIN_VALUE, Integer.MAX_VALUE, 
                            Math.max(left.maxSize, right.maxSize));
    }
    
    public int largestBSTSubtree(TreeNode root) {
        return largestBSTSubtreeHelper(root).maxSize;
    }
}