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Data Structures | Balanced Binary Search Trees | Question 2

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Discuss

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 logarithmic 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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Last Updated : 28 Jun, 2021

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