Explanation : When we are sorting an array of n integers, Recurrence relation for Total number of comparisons involved will be,
T(n) = 2T(n/2) + (n) where (n) is the number of comparisons in order to merge 2 sorted subarrays of size n/2.
Instead of integers whose comparison take O(1) time, we are given n strings. We can compare 2 strings in O(n) worst case. Therefore, the total number of comparisons now will be (n2log2n) where each comparison takes O(n) time now.
In general, merge sort makes (nlog2n) comparisons, and runs in (nlog2n) time if each comparison can be done in O(1) time.
Explanation : Since Inorder traversal of a BST always gives elements in increasing order. For this question sorted order of keys will always be the in-order tree traversal for post-order traversal. So, option (D) is correct.