The leaves of \(T\) are leaves of initial stars, and those vertices distant by at most two from it belong to the same star. Thus, the following algorithm is possible.

• Choose a leaf of \(T\).
• Count the number of vertices distant by at most two from it (including itself). If there are \(x\) of them, they form a level-\((x-1)\) star.
• Remove the counted vertices and adjacent edges from \(T\).

After you repeat this until no vertices remain in \(T\), you obtain the answer. With an appropriate implementation, it works in a total of \(\mathrm{O}(N)\) time, but the implementation is a bit complicated. We describe a simpler implementation with some observations.

The following lemma holds.

• In an original star, let us call the non-leaf vertex the center. In \(T\), the distance between centers is always a multiple of \(3\), and that between a center and a leaf is always a non-multiple of \(3\).

This can be shown by induction. With this lemma, we can come up with the following algorithm:

• Choose a vertex from \(T\) that was the center of an original star. You can do so by choosing the vertex adjacent to a leaf.
• Find the shortest distance of each vertex from the chosen vertex.
• For each vertex whose shortest distance is a multiple of \(3\), add the degree of that vertex to \(L\).

This algorithm works in a total of \(\mathrm{O}(N)\) times, and requires a simple algorithm like BFS (Breadth-First Search) or DFS (Depth-First Search).