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 -
rootis 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
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