700-search-in-a-binary-search-tree

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Description

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(Python)

# 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]:
        return self.iterative(root, val)

    # Time Complexity: O(log n)
    # Space Complexity: O(log n)
    def recursive(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
        if root is None:
            return None
        if root.val == val:
            return root
        elif root.val > val:
            return self.searchBST(root.left, val)
        elif root.val < val:
            return self.searchBST(root.right, val)

    # Time Complexity: O(log n)
    # Space Complexity: O(1)
    def iterative(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
        cur = root

        while cur is not None:
            if cur.val == val:
                return cur
            elif cur.val > val:
                cur = cur.left
            else:
                cur = cur.right

        return cur