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.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isValidBST(TreeNode root) {
        if (root == null)
            return true;
        if (root.left != null && !smallerThanRoot(root, root.left))
            return false;
        if (root.right != null && !greaterThanRoot(root, root.right))
            return false;
        if (isValidBST(root.left) && isValidBST(root.right))
            return true;
        return false;
    }
    
    private boolean greaterThanRoot(TreeNode root, TreeNode child) {
        if (child.val <= root.val)
            return false;
        if (child.left != null) {
            if (!greaterThanRoot(root, child.left))
                return false;
        }
        if (child.right != null) {
            if (!greaterThanRoot(root, child.right))
                return false;
        }
        return true;
    }
    
    private boolean smallerThanRoot(TreeNode root, TreeNode child) {
        if (child.val >= root.val)
            return false;
        if (child.left != null) {
            if (!smallerThanRoot(root, child.left))
                return false;
        }
        if (child.right != null) {
            if (!smallerThanRoot(root, child.right))
                return false;
        }
        return true;
    }
}

Solution 2 -- Inorder Traversal

Inorder traversal of BST is an ascending array. Java Stack

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isValidBST(TreeNode root) {
        // This problem can be looked as inorder traversal problem
        // Inorder traversal of BST is an ascending array
        List<Integer> inOrderResult = new ArrayList<Integer>();
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode tmp = root;
        while (tmp != null || !stack.empty()) {
            if (tmp != null) {
                stack.push(tmp);
                tmp = tmp.left;
            } else {
                TreeNode current = stack.pop();
                inOrderResult.add(current.val);
                tmp = current.right;
            }
        }
        // Traverse list
        if (inOrderResult.size() < 1)
            return true;
        int max = inOrderResult.get(0);
        for (int i = 1; i < inOrderResult.size(); i++) {
            if (inOrderResult.get(i) > max)
                max = inOrderResult.get(i);
            else
                return false;
        }
        return true;
    }
}