Alternate Splay

In our implementation of the insert operation for splay trees, when insert(S, x) is called, we first call splay(S, x). Supposing that x is not already in S, the splay operation brings up either the predecessor or the successor of x to the root. Then, we make x either the left or right child of this root. There is an alternate implementation of insert. First, insert x as you would normally do in a binary search tree. Then, rotate x as you would in the splay operation until x is the root of the tree. These two implementations are actually different. In the original implementation, x does not become the root of the tree (unless the tree was empty originally). In the alternate implementation, x does become the root. Describe how the amortized analysis of the splay operation can be modified to show that the alternate implementation of the insert operation also takes O(log n) amortized time. (You only have to explain how to modify the amortized analysis. Do not reproduce the entire proof.)