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Data Structures | Balanced Binary Search Trees | Question 2

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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 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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