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Data Structures | Balanced Binary Search Trees | Question 2

Last Updated : 28 Jun, 2021

The worst case running time to search for an element in a balanced in a binary search tree with n2^n elements is

(A) $\Theta(n log n)$
(B) $\Theta (n2^n)$
(C) $\Theta (n)$
(D) $\Theta (log n)$

(A) A
(B) B
(C) C
(D) D

Answer: (C)

Explanation: Time taken to search an element is $\Theta (h)$ where h is the height of Binary Search Tree (BST). The growth of height of a balanced BST is logerthimic in terms of number of nodes. So the worst case time to search an element would be $\Theta (Log(n*2^n))$ which is $\Theta (Log(n) + Log(2^n))$ Which is $\Theta (Log(n) + n)$ which can be written as $\Theta (n)$ .

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