Question

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

Answer 1

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 Θ(Log(n*2^n)) which is Θ(Log(n) + Log(2^n)) Which is Θ(Log(n) + n) which can be written as Θ(n).

Answer 2

Explanation:

the worst case running time to search for an element in a balanced in a binary search tree with