## DEV Community

Abhishek Chaudhary

Posted on

# Search in a Binary Search Tree

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`

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]:
if root:
if root.val == val:
return root
elif val > root.val:
return self.searchBST(root.right, val)
else:
return self.searchBST(root.left, val)
else:
return None
``````