Example of tree with > 6 vertices, tree would have depth = n after splay() deepest vertex

Question

How to build tree with more than 6 vertices, that after operation splay() would have depth = number of vertices? Is it possible?

UPD: Example for n = 4:

• insert 60
• insert 10
• insert 20
• insert 50
• splay(10)

You can use splay tree visualization

Answer 1

This is a good question whose answer can help us understand how splay tree tends to keep away from being imbalanced.

No, it is impossible for any rooted tree with more than 5 vertices to become a linear tree after operation splay() one of the deepest vertices, i.e. its depth is one less than the number of vertices.

Why?

For the sake of contradiction, let there be a rooted tree with more than 5 vertices which becomes a linear tree after splaying $x$, one of its deepest vertices. Consider the point of time just before the last splay step that moves $x$ to the root. There are three cases.

• That last step is a zig-step (the following graph) or its mirroring zag-step. Since the resulted tree is linear, part $A$ and $B$ are empty. That means, the left subtree of the root before that step contains node $x$ only. Since $x$ is the deepest node, the whole tree has at most 3 nodes, which is not true.

• That last step is zig-zig step (the following graph) or its mirroring zag-zag. Since the resulted tree is linear, part $A$, $B$ and $C$ are empty. That means, the left subtree of the root before that step contains node $P$ and $x$ only. Part $D$ can have at most two nodes; otherwise, either $x$ was not the deepest node or the resulted tree is not linear. So the whole tree has at most 5 nodes, which is not true.

• That last step is a zig-zag step (the following graph) or its mirroring zag-zig. However, after that step, the root will alway has two children, which is not ture.

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Exercise. Draw a tree with 5 vertices that becomes a linear tree after splaying one of its deepest vertices.