700. Search in a Binary Search Tree

Easy LeetCode

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

How to solve the problem

Code

• Recursion (DFS) For BST

# 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]:
        # For BST
        if not root:
            return None
        if root.val == val:
            return root
        if val < root.val:
            return self.searchBST(root.left, val)
        else:
            return self.searchBST(root.right, val)

• Recursion (DFS) For General Binary 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]:
        # For general Binary Tree
        if not root:
            return None
        if root.val == val:
            return root
        return self.searchBST(root.left, val) or self.searchBST(root.right, val)

Complexity

• Time complexity: O(n), n == number of nodes
• Space complexity: O(n)