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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:

The Largest BST Subtree in this case is the highlighted one.
The return value is the subtree's size, which is 3.

Solution:

 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 class Result {
12         int min;
13         int max;
14         boolean isBST;
15         int maxBSTNodeNum;
16         public Result(int i, int x, boolean bst, int num){
17             min = i;
18             max = x;
19             isBST = bst;
20             maxBSTNodeNum = num;
21         }
22     }
23     public int largestBSTSubtree(TreeNode root) {
24         if (root==null) return 0;
25         
26         Result res = largestBSTSubtreeRecur(root);
27         return res.maxBSTNodeNum;
28         
29     }
30     
31     public Result largestBSTSubtreeRecur(TreeNode cur){
32         boolean isBST = true;
33         int min = cur.val;
34         int max = cur.val;
35         int nodeNum = 1;
36         int subBSTNodeNum = 0;
37         
38         Result leftRes=null, rightRes=null;
39         if (cur.left!=null){
40             leftRes = largestBSTSubtreeRecur(cur.left);
41             if (!leftRes.isBST) isBST = false;
42             else {
43                 if (leftRes.max>=cur.val) isBST = false;
44                 else {
45                     min = leftRes.min;
46                 }
47             }
48             if (isBST) subBSTNodeNum += leftRes.maxBSTNodeNum;
49         }
50         
51         if (cur.right!=null){
52             rightRes = largestBSTSubtreeRecur(cur.right);
53             if (!rightRes.isBST) isBST = false;
54             else {
55                 if (rightRes.min<=cur.val) isBST = false;
56                 else {
57                     max = rightRes.max;
58                 }
59             }
60             if (isBST) subBSTNodeNum += rightRes.maxBSTNodeNum;
61         }
62         
63         if (isBST) nodeNum = subBSTNodeNum+1;
64         else {
65             if (leftRes!=null) nodeNum = Math.max(nodeNum,leftRes.maxBSTNodeNum);
66             if (rightRes!=null) nodeNum = Math.max(nodeNum,rightRes.maxBSTNodeNum);
67         }
68         
69         return new Result(min,max,isBST,nodeNum);
70     }
71 }

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原文地址：https://www.cnblogs.com/lishiblog/p/5741330.html