Leetcode: Largest BST Subtree

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.
Here's an example:
    10
    / 
   5  15
  /     
 1   8   7
The Largest BST Subtree in this case is the highlighted one. 
The return value is the subtree's size, which is 3.
Hint:

You can recursively use algorithm similar to 98. Validate Binary Search Tree at each node of the tree, which will result in O(nlogn) time complexity.
Follow up:
Can you figure out ways to solve it with O(n) time complexity?

refer to https://discuss.leetcode.com/topic/36995/share-my-o-n-java-code-with-brief-explanation-and-comments/2

这道题不好从root到leaf一层一层限制subtree取值范围，因为有可能parent并不能构成BST，反倒是需要如果subtree是BST的话，限制parent的取值范围，然后根据情况判断是：

1. 吸纳parent以及parent的另一个subtree进来形成一个更大的BST，向上传递这个新subtree的size

2. parent向上传递自它以下发现的最大BST

所以我们需要传递subtree size, subtree (min, max)范围，所以我们想到用一个wrapper class包括这三项东西。

同时因为要能分辨上面1、2两种情况，所以size为正表示1，size为负表示2

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public class Result {
        int res;
        int min;
        int max;
        public Result(int num, int n1, int n2) {
            this.res = num;
            this.min = n1;
            this.max = n2;
        }
    }
    
    public int largestBSTSubtree(TreeNode root) {
        Result res = findLargestBST(root);
        return Math.abs(res.res);
    }
    
    public Result findLargestBST(TreeNode cur) {
        if (cur == null) return new Result(0, Integer.MAX_VALUE, Integer.MIN_VALUE);
        Result left = findLargestBST(cur.left);
        Result right = findLargestBST(cur.right);
        if (left.res<0 || right.res<0 || cur.val<left.max || cur.val>right.min) {
            return new Result(Math.max(Math.abs(left.res), Math.abs(right.res))*(-1), 0, 0);
        }
        else return new Result(left.res+right.res+1, Math.min(left.min, cur.val), Math.max(right.max, cur.val));
    }
}