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




 # 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 not root:
            return

        if root.val == val:
            return root

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

Random Note

From python 3.7 dict guarantees that order will be kept as they inserted, and popitem will use LIFO order but we need FIFO type system. so we need OrderedDict which have popIten(last = T/F) for this req. One thing, next(iter(dict)) will return the first key of the dict

Python