LeetCode - Easy - 700. Search in a Binary Search Tree

Topic

• Tree

Description

https://leetcode.com/problems/search-in-a-binary-search-tree/

You are given the root of a binary search tree (BST) and an integer val.

Find the node in the BST that the node’s value equals val and return the subtree rooted with that node. If such a node does not exist, return null.

Example 1:

Input: root = [4,2,7,1,3], val = 2
Output: [2,1,3]

Example 2:

Input: root = [4,2,7,1,3], val = 5
Output: []

Constraints:

• The number of nodes in the tree is in the range $[1,5000]$.
• $1<=Node.val<=10^7$
• root is a binary search tree.
• $1<=val<=10^7$

Analysis

方法一：递归法

方法二：迭代法

Submission

import com.lun.util.BinaryTree.TreeNode;

public class SearchInABinarySearchTree {
	
	//方法一：递归法
    public TreeNode searchBST(TreeNode root, int val) {
        if(root == null) return null;
    	
    	if(val < root.val)
    		return searchBST(root.left, val);
    	else if(root.val < val)
    		return searchBST(root.right, val);
    	else 
    		return root;
    }
    
    //方法二：迭代法
    public TreeNode searchBST2(TreeNode root, int val) {
    	TreeNode p = root;
    	
    	while(p != null) {
    		if(val < p.val) {
    			p = p.left;
    		}else if(val > p.val){
    			p = p.right;
    		}else {
    			return p;
    		}
    	}
    	
    	return null;
    } 
    
}

Test

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

import com.lun.util.BinaryTree;
import com.lun.util.BinaryTree.TreeNode;

public class SearchInABinarySearchTreeTest {

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

		TreeNode root = BinaryTree.integers2BinaryTree(4,2,7,1,3);
		TreeNode expected = BinaryTree.integers2BinaryTree(2,1,3);
		
		assertTrue(BinaryTree.equals(obj.searchBST(root, 2), expected));
		assertNull(obj.searchBST(root, 5));
	}
	
	@Test
	public void test2() {
		SearchInABinarySearchTree obj = new SearchInABinarySearchTree();
		
		TreeNode root = BinaryTree.integers2BinaryTree(4,2,7,1,3);
		TreeNode expected = BinaryTree.integers2BinaryTree(2,1,3);
		
		assertTrue(BinaryTree.equals(obj.searchBST2(root, 2), expected));
		assertNull(obj.searchBST2(root, 5));
	}
}