Validate Binary Search Tree

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.

 

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1

  / \

 2   3

    /

   4

    \

     5

The above binary tree is serialized as  "{1,2,3,#,#,4,#,#,5}".

 

BST如果进行inorder 的话应该是sorted。

 1 /**

 2  * Definition for binary tree

 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     int per = 0;

12     boolean sig;

13     public boolean isValidBST(TreeNode root) {

14         // IMPORTANT: Please reset any member data you declared, as

15         // the same Solution instance will be reused for each test case.

16         per = Integer.MIN_VALUE;

17         sig = true;

18         isValid(root);

19         return sig;

20     }

21     public void isValid(TreeNode root){

22         if(root == null) return;

23         isValid(root.left);

24         if(root.val > per) per = root.val;

25         else{

26             sig = false;

27             return;

28         }

29         isValid(root.right);

30     }

31 }

还是可以使用递归来做：

 1 public class Solution {

 2     public boolean isValidBST(TreeNode root) {

 3         // IMPORTANT: Please reset any member data you declared, as

 4         // the same Solution instance will be reused for each test case.

 5         return root == null || check(root.left, Integer.MIN_VALUE, root.val) && check(root.right, root.val, Integer.MAX_VALUE);

 6     }

 7     public boolean check(TreeNode root, int min, int max){

 8         if(root == null) return true;

 9         if(!(min < root.val && root.val < max)) return false;

10         return check(root.left, min, root.val) && check(root.right, root.val, max);

11     }

12 }

这个方法很好！