Description
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
.
Intuition
There are only two possibilities when we searching a node:
- node is equal to the value
- node is not equal to the value
Algorithm
When the node is equal to the value, we just need to return it. Otherwise, we just need to visit its left or right child recursively.
Pseudocode
visit root node.
If root is null or root’s value is equal to target value, return the root.
Otherwises, if root’s value is bigger than target value.
visit root’s left child.
Otherwises
visit root’s right child.
Code
# 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 == None or root.val == val):
return root
elif root.val < val:
return self.searchBST(root.right, val)
else:
return self.searchBST(root.left, val)
Complexity Analysis
- Time Complexity: $O(\log{n})$
- Space Complexity: $O(1)$