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

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The worst case running time to search for an element in a balanced binary search tree with n2n elements is 

(A) ⊖ (n log n) 

(B) ⊖(n2n)

(C) ⊖(n) 

(D) ⊖(log n)

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(C) ⊖(n) 

The worst case search t ime in a balanced BST on 'x' nodes is log x . So, if x = n2n', 

then log(n2n) = logn + log(2n) = log n + n = 8( n)

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