LeetCode - Medium - 98. Validate Binary Search Tree

Topic

• Tree
• Recursion
• Depth-first Search

Description

https://leetcode.com/problems/validate-binary-search-tree/

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

A valid 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.

Example 1:

Input: root = [2,1,3]
Output: true

Example 2:

Input: root = [5,1,4,null,null,3,6]
Output: false
Explanation: The root node's value is 5 but its right child's value is 4.

Constraints:

• The number of nodes in the tree is in the range $[1,10^4]$.
• $-2^{31}<=Node.val<=231-1$

Analysis

方法一：我写的，中序遍历模式递归版。

方法二：我写的，中序遍历模式迭代版。

方法三：别人写的，前序遍历模式递归版。

Submission

import java.util.LinkedList;

import com.lun.util.BinaryTree.TreeNode;

public class ValidateBinarySearchTree {
	
	//方法一：我写的，中序遍历模式递归版
    public boolean isValidBST(TreeNode root) {
    	Integer[] prev = {null};
    	boolean[] result = {true};
    	isValidBST(root, prev, result);
    	return result[0];
    }
    
    private void isValidBST(TreeNode node, Integer[] prev, boolean[] result) {
    	if(node == null || !result[0]) return;
    	isValidBST(node.left, prev, result);
    	if(prev[0] != null && prev[0] >= node.val) {
    		result[0] = false;
    		return;
    	}
    	prev[0] = node.val;
    	isValidBST(node.right, prev, result);
    }
    
    //方法二：我写的，中序遍历模式迭代版
    public boolean isValidBST2(TreeNode root) {
    	LinkedList<TreeNode> stack = new LinkedList<>();
    	TreeNode p = root;
    	Integer prev = null;
    	while(!stack.isEmpty() || p != null) {
    		if(p != null) {
    			stack.push(p);
    			p = p.left;
    		}else {
    			TreeNode node = stack.pop();
    			if(prev != null && prev >= node.val)
    				return false;
    			prev = node.val;
    			p = node.right;
    		}
    	}
    	return true;
    }
    
    //方法三：别人写的，前序遍历模式递归版
    public boolean isValidBST3(TreeNode root) {
        return isValidBST3(root, null, null);
    }

    private boolean isValidBST3(TreeNode root, Integer min, Integer max) {
        if(root == null) return true;
        if(min != null && root.val <= min) return false;
        if(max != null && root.val >= max) return false;
        
        return isValidBST3(root.left, min, root.val) && isValidBST3(root.right, root.val, max);
    }
    
}

Test

import static org.junit.Assert.*;
import org.junit.Test;

import com.lun.util.BinaryTree;

public class ValidateBinarySearchTreeTest {

	@Test
	public void test() {
		ValidateBinarySearchTree obj = new ValidateBinarySearchTree();

		assertTrue(obj.isValidBST(BinaryTree.integers2BinaryTree(2,1,3)));
		assertFalse(obj.isValidBST(BinaryTree.integers2BinaryTree(5,1,4,null,null,3,6)));
		assertFalse(obj.isValidBST(BinaryTree.integers2BinaryTree(5,4,6,null,null,3,7)));
	}
	
	@Test
	public void test2() {
		ValidateBinarySearchTree obj = new ValidateBinarySearchTree();
		
		assertTrue(obj.isValidBST2(BinaryTree.integers2BinaryTree(2,1,3)));
		assertFalse(obj.isValidBST2(BinaryTree.integers2BinaryTree(5,1,4,null,null,3,6)));
		assertFalse(obj.isValidBST2(BinaryTree.integers2BinaryTree(5,4,6,null,null,3,7)));
	}
	
	@Test
	public void test3() {
		ValidateBinarySearchTree obj = new ValidateBinarySearchTree();
		
		assertTrue(obj.isValidBST2(BinaryTree.integers2BinaryTree(2,1,3)));
		assertFalse(obj.isValidBST2(BinaryTree.integers2BinaryTree(5,1,4,null,null,3,6)));
		assertFalse(obj.isValidBST2(BinaryTree.integers2BinaryTree(5,4,6,null,null,3,7)));
	}
}