# LeetCode 700. Search in a Binary Search Tree

## Description

https://leetcode.com/problems/swap-nodes-in-pairs/

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`

## Explanation

Search recursively base on binary search tree characteristics.

## Python 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: TreeNode, val: int) -> TreeNode:
return self.helper(root, val)

def helper(self, root, val):
if not root:
return None

if root.val == val:
return root

if val < root.val:
return self.helper(root.left, val)
else:
return self.helper(root.right, val)

return None
``````
• Time Complexity: O(H). H is the tree height.
• Space Complexity: O(H).