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:

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public int largestBSTSubtree(TreeNode root) {
12         int [] res = {0};
13         helper(root, res);
14         return res[0];
15     }
16     
17     private Node helper(TreeNode root, int [] res){
18         Node cur = new Node();
19         if(root == null){
20             cur.isBST = true;
21             return cur;
22         }
23         Node left = helper(root.left, res);
24         Node right = helper(root.right, res);
25         if(left.isBST && root.val > left.max && right.isBST && root.val < right.min){
26             cur.isBST = true;
27             cur.min = Math.min(root.val, left.min);
28             cur.max = Math.max(root.val, right.max);
29             cur.size = left.size + right.size + 1;
30             if(cur.size > res[0]){
31                 res[0] = cur.size;
32             }
33         }
34         return cur;
35     }
36 }
37 
38 class Node{
39     boolean isBST;
40     int min;
41     int max;
42     int size;
43     public Node(){
44         isBST = false;
45         min = Integer.MAX_VALUE;
46         max = Integer.MIN_VALUE;
47         size = 0;
48     }
49 }