1) The time complexity is always O(n.logn) and hence, this statement is wrong.
2) COnsider a skewed tree. Now, if you want a complete balanced tree, the no of nodes in the bottommost level is approx $\frac{n}{2}$
Each rotation will bring 1 node to the bottommost level . So, I think it will be O(n/2) = O(n)
3) Computing all the node values results in extremely bad time complexity. DP and memoization(not memorization) results in O(no of nodes in the tree) time complexity.
Memoization is remembering previously computed values so that we dont do the same job again.
So, as per me, D is the right answer.