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While the remaining root has a left child, apply a right tree rotation there. Once the left child is null, output the root and advance to its right child. The output already produced followed by the remaining inorder sequence is invariant. There are exactly $N$ output removals and $L(T_0)\le N$ rotations, because rotations decrease the left-size potential and removals preserve it. The algorithm therefore takes linear time and constant auxiliary pointer space, at the cost of changing the input links.

Define $L(\varnothing )=0$ and $L(T)=|T.left|+L(T.right)$. Subtracting the right-spine length from total size gives the same quantity, because every node lies either on that spine or in one of its disjoint left subtrees. A right tree rotation at the root decreases this potential by exactly one; removing a root with no left child preserves it. Thus repeated root rotations and inorder removals take linear total work.