Question: Consider the following simpler alternative to splaying: MoveToRoot(v): while parent(v) != null single rotate at v Prove that the amortized cost of MoveToRoot in an

Consider the following simpler alternative to splaying: MoveToRoot(v): while parent(v) != null single rotate at v Prove that the amortized cost of MoveToRoot in an n-node binary tree can be (n). That is, prove that for any integer k, there is a sequence of k MoveToRoot operations that require (kn) time to execute.

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