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, 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

    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