GATE | GATE CS 2012 | Question 5
The worst case running time to search for an element in a balanced in a binary search tree with n2^n elements is
(A)
(B)
(C)
(D)
(A) A
(B) B
(C) C
(D) D
Answer: (C) Explanation:
-> The search time in a binary search tree depends on the form of the tree, that is on the order in which its nodes were inserted. A pathological case: The n nodes were inserted by increasing order of the keys, yielding something like a linear list (but with a worse space consumption), with O(n) search time(in the case of skew tree).
-> A balanced tree is a tree where every leaf is “not more than a certain distance” away from the root than any other leaf.So in balanced tree, the height of the tree is balanced to make distance between root and leafs nodes a low as possible. In a balanced tree, the height of tree is log
2(n).
-> So , if a Balanced Binary Search Tree contains n2n elements then Time complexity to search an item:
Time Complexity = log(n2
n) = log (n) + log(2
n)
= log (n) +n = O(n)
So Answer is C.
See
https://www.geeksforgeeks.org/data-structures-and-algorithms-set-28/
This solution is contributed by
Nirmal BharadwajQuiz of this Question
Last Updated :
28 Jun, 2021
Like Article
Save Article
Share your thoughts in the comments
Please Login to comment...