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).