700.Search_in_a_Binary_Search_Tree(Easy).md

700. Search in a Binary Search Tree (Easy)

Date and Time: Jul 23, 2024, 18:27 (EST)

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

Question:

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

KeyPoints:

Follow the property of BST and update cur to be its left or right depends on cur.val and val.

My Solution:

# 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]:
        cur = root
        while cur:
            if cur.val < val:
                cur = cur.right
            elif cur.val > val:
                cur = cur.left
            else:
                return cur
        return None

Time Complexity: $O(log\ n)$
Space Complexity: $O(1)$