# Solving Leetcode Interviews in Seconds with AI: Search in a Binary Search Tree


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "700" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# 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  

	# Explanation
	Here's an efficient solution to find a node in a BST with a specific value:

*   **BST Property:** Leverage the inherent property of Binary Search Trees. If the target value is less than the current node's value, search the left subtree; if it's greater, search the right subtree.
*   **Iterative Approach:** Implement an iterative approach for better space efficiency (avoiding recursive call stack).
*   **Early Exit:** Return `None` immediately if the root is `None`.

*   **Runtime Complexity:** O(log n) on average, O(n) in the worst case (skewed tree). **Storage Complexity:** O(1).

	
	# Code
	```python
	class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def searchBST(root, val):
    """
    Finds a node in a BST with a specific value and returns the subtree rooted with that node.

    Args:
        root: The root node of the BST.
        val: The value to search for.

    Returns:
        The subtree rooted with the node whose value equals val, or None if such a node does not exist.
    """
    while root:
        if val < root.val:
            root = root.left
        elif val > root.val:
            root = root.right
        else:
            return root
    return None
	```