Yes, I agree. In fact we can take the simple example here. T1's root is 2, left of root 1 and right of root 3. Let T2's root be 3, left of root 1 and right of 1 be 2. There's no way we can get T2 from T1 using right rotations. But if given that T1 with 'n' nodes can be converted to T2, in the worst case how many such right-rotations are needed. The answer is supposed to be O(n^2) but I couldn't prove it.
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