98. Validate Binary Search Tree

Posted 2020-07-12 烁宝宝

Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:

• The left subtree of a node contains only nodes with keys less than the node‘s key.
• The right subtree of a node contains only nodes with keys greater than the node‘s key.
• Both the left and right subtrees must also be binary search trees.

思路：中序排列为排序好的数组（数组元素一次增大）；数组中不能有重复元素。

代码如下：

 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 boolean isValidBST(TreeNode root) {
12       
13         if(root==null||(root.left==null&&root.right==null))
14         return true;
15         ArrayList<Integer> list=ParentAndSon(root);
16           int[] nums=new int[list.size()];
17           int[] nums1=new int[list.size()];
18           for(int i=0;i<list.size();i++)
19           {
20               nums[i]=list.get(i);
21               nums1[i]=list.get(i);
22           }
23           Arrays.sort(nums);
24     
25           if(nums[0]!=nums1[0])
26           return false;
27           for(int i=1;i<nums.length;i++)
28           {
29            if(nums1[i]!=nums[i]||nums[i]==nums[i-1])
30            return false;
31           
32           }
33      return true;
34     }
35     public ArrayList<Integer> ParentAndSon(TreeNode root){
36         ArrayList<Integer> list=new ArrayList<>();
37         if(root.left!=null)
38         list.addAll(ParentAndSon(root.left));
39         
40         list.add(root.val);
41         
42         if(root.right!=null)
43         list.addAll(ParentAndSon(root.right));
44         return list;
45     }
46 }