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Data Structures | Balanced Binary Search Trees | Question 2

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  • Last Updated : 28 Jun, 2021
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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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