Validate Binary Search Tree 解答

Question

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.

Solution 1 -- Recursive

According to the question, we can write recursive statements. Note here whole left/right subtree should be smaller/greater than the root.

 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         if (root == null)
13             return true;
14         if (root.left != null && !smallerThanRoot(root, root.left))
15             return false;
16         if (root.right != null && !greaterThanRoot(root, root.right))
17             return false;
18         if (isValidBST(root.left) && isValidBST(root.right))
19             return true;
20         return false;
21     }
22
23     private boolean greaterThanRoot(TreeNode root, TreeNode child) {
24         if (child.val <= root.val)
25             return false;
26         if (child.left != null) {
27             if (!greaterThanRoot(root, child.left))
28                 return false;
29         }
30         if (child.right != null) {
31             if (!greaterThanRoot(root, child.right))
32                 return false;
33         }
34         return true;
35     }
36
37     private boolean smallerThanRoot(TreeNode root, TreeNode child) {
38         if (child.val >= root.val)
39             return false;
40         if (child.left != null) {
41             if (!smallerThanRoot(root, child.left))
42                 return false;
43         }
44         if (child.right != null) {
45             if (!smallerThanRoot(root, child.right))
46                 return false;
47         }
48         return true;
49     }
50 }

Solution 2 -- Inorder Traversal

Inorder traversal of BST is an ascending array. Java Stack

 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         // This problem can be looked as inorder traversal problem
13         // Inorder traversal of BST is an ascending array
14         List<Integer> inOrderResult = new ArrayList<Integer>();
15         Stack<TreeNode> stack = new Stack<TreeNode>();
16         TreeNode tmp = root;
17         while (tmp != null || !stack.empty()) {
18             if (tmp != null) {
19                 stack.push(tmp);
20                 tmp = tmp.left;
21             } else {
22                 TreeNode current = stack.pop();
23                 inOrderResult.add(current.val);
24                 tmp = current.right;
25             }
26         }
27         // Traverse list
28         if (inOrderResult.size() < 1)
29             return true;
30         int max = inOrderResult.get(0);
31         for (int i = 1; i < inOrderResult.size(); i++) {
32             if (inOrderResult.get(i) > max)
33                 max = inOrderResult.get(i);
34             else
35                 return false;
36         }
37         return true;
38     }
39 }