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.

一刷
题解：
如果root比左孩子要大是不够的，还要比左孩子的subtree的最小值要大。于是考虑创建一个object来保存。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    
    class Result{
        int size;
        int lower;
        int upper;
        // size of current tree, range of current tree [rangeLower, rangeUpper]
        
        Result(int size, int lower, int upper){
            this.size = size;
            this.lower = lower;
            this.upper = upper;
        }
    }
    
    int max = 0;    
    
    public int largestBSTSubtree(TreeNode root) {
        if(root == null) return 0;
        traverse(root);
        return max;
    }
    
    private Result traverse(TreeNode root){
        if(root == null) return new Result(0, Integer.MAX_VALUE, Integer.MIN_VALUE);
        Result left = traverse(root.left);
        Result right = traverse(root.right);
        if(left.size == -1 || right.size == -1 || root.val <= left.upper || root.val >= right.lower)
                return new Result(-1, 0, 0);
        int size = left.size + 1 + right.size;
        max = Math.max(size, max);
        return new Result(size, Math.min(left.lower, root.val), Math.max(right.upper, root.val));
    }
  
}

二刷
题意是给一个binary tree, 问里面满足binary search tree的subtree的节点数目是多少。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    class Subtree{
        int lower;
        int upper;
        int size;
        
        Subtree(int lower, int upper, int size){
            this.lower = lower;
            this.upper = upper;
            this.size = size;
        }
    }
    int max = 0;
    public int largestBSTSubtree(TreeNode root) {
        if(root == null) return 0;
        traverse(root);
        return max;
    }
    
    private Subtree traverse(TreeNode root){
        if(root == null){
            return new Subtree(Integer.MAX_VALUE, Integer.MIN_VALUE, 0);
        }
        Subtree left = traverse(root.left);
        Subtree right = traverse(root.right);
        int size = 0;
        if(left.size!= -1 && right.size!=-1 && root.val > left.upper && root.val<right.lower) size = left.size + 1 + right.size;
        else return new Subtree(0, 0, -1);
        max = Math.max(size, max);
        return new Subtree(Math.min(left.lower, root.val), Math.max(right.upper, root.val), size);//narrow the range
    }
}