LeetCode #700 — EASY

Search in a Binary Search Tree

Build confidence with an intuition-first walkthrough focused on tree fundamentals.

The Problem

Problem Statement

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 <= 10⁷
root is a binary search tree.
1 <= val <= 10⁷

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Tree

Example 1

[4,2,7,1,3]
2

Example 2

[4,2,7,1,3]
5

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #700: Search in a Binary Search Tree
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode searchBST(TreeNode root, int val) {
        if (root == null || root.val == val) {
            return root;
        }
        return root.val > val ? searchBST(root.left, val) : searchBST(root.right, val);
    }
}

// Accepted solution for LeetCode #700: Search in a Binary Search Tree
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
 func searchBST(root *TreeNode, val int) *TreeNode {
    if root == nil || root.Val == val {
        return root
    }
    if root.Val > val {
        return searchBST(root.Left, val)
    }
    return searchBST(root.Right, val)
}

# Accepted solution for LeetCode #700: Search in a Binary Search Tree
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
        if root is None or root.val == val:
            return root
        return (
            self.searchBST(root.left, val)
            if root.val > val
            else self.searchBST(root.right, val)
        )

// Accepted solution for LeetCode #700: Search in a Binary Search Tree
struct Solution;
use rustgym_util::*;

trait Search {
    fn find(&self, val: i32) -> TreeLink;
}

impl Search for TreeLink {
    fn find(&self, val: i32) -> TreeLink {
        if let Some(node) = self {
            let temp = node.clone();
            let node = node.borrow();
            if val == node.val {
                Some(temp)
            } else {
                if val < node.val {
                    Self::find(&node.left, val)
                } else {
                    Self::find(&node.right, val)
                }
            }
        } else {
            None
        }
    }
}

impl Solution {
    fn search_bst(root: TreeLink, val: i32) -> TreeLink {
        root.find(val)
    }
}

#[test]
fn test() {
    let root = tree!(4, tree!(2, tree!(1), tree!(3)), tree!(3));
    let res = tree!(2, tree!(1), tree!(3));
    assert_eq!(Solution::search_bst(root, 2), res);
}

// Accepted solution for LeetCode #700: Search in a Binary Search Tree
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function searchBST(root: TreeNode | null, val: number): TreeNode | null {
    if (root === null || root.val === val) {
        return root;
    }
    return root.val > val ? searchBST(root.left, val) : searchBST(root.right, val);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.