LeetCode 333. Largest BST Subtree

原题链接在这里：https://leetcode.com/problems/largest-bst-subtree/

题目：

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.

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

题解：

采用bottom-up的方法，简历新的class, 用来存储

• 当前节点为root的subtree是否是BST
• 若是，最小val 和最大val.
• size是当前subtree的大小.

然后从下到上更新，若是中间过程中size 比 res大，就更新res.

Time Complexity: O(n). 每个点不会访问超过两遍. Space: O(logn). Recursion stack space.

AC Java:

 /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public int largestBSTSubtree(TreeNode root) {
         int [] res = {0};
         helper(root, res);
         return res[0];
     }

     private Node helper(TreeNode root, int [] res){
         Node cur = new Node();
         if(root == null){
             cur.isBST = true;
             return cur;
         }
         Node left = helper(root.left, res);
         Node right = helper(root.right, res);
         if(left.isBST && root.val > left.max && right.isBST && root.val < right.min){
             cur.isBST = true;
             cur.min = Math.min(root.val, left.min);
             cur.max = Math.max(root.val, right.max);
             cur.size = left.size + right.size + 1;
             if(cur.size > res[0]){
                 res[0] = cur.size;
             }
         }
         return cur;
     }
 }

 class Node{
     boolean isBST;
     int min;
     int max;
     int size;
     public Node(){
         isBST = false;
         min = Integer.MAX_VALUE;
         max = Integer.MIN_VALUE;
         size = 0;
     }
 }