Keep The Heavy Child

DSU on tree answers one subtree query per node in $O(n \log n)$ . What is the counting argument behind the log?

Solve the light children first and clear their contribution, keep the heavy child's state, then re-add the light subtrees. Any root-to-node path crosses $\Theta(\sqrt n)$ light edges, so each node is added $\Theta(\sqrt n)$ times overall.

Solve the light children first and clear their contribution, keep the heavy child's state, then re-add the light subtrees. Any root-to-node path crosses at most $\log n$ light edges, so each node is added $O(\log n)$ times overall.

Solve the light children first and clear their contribution, keep the heavy child's state, then re-add the light subtrees. Any root-to-node path has depth at most $\log n$ after decomposition, so each node is added $O(\log n)$ times overall.

Solve the heavy child first and clear its contribution, keep the light children's state, then re-add the heavy subtree. Any root-to-node path crosses at most $\log n$ light edges, so each node is added $O(\log n)$ times overall.