Complexity Analysis (Linear time)

Both DFS passes visit each node exactly once. The down pass computes size[v] and down[v] for all $n$ nodes. The up pass computes answer[v] for all $n$ nodes. Each pass is $O(n)$ time.

Total time: $O(n)$ for building the adjacency list, $O(n)$ for the down pass, $O(n)$ for the up pass. That gives $O(n)$ overall.

Space: $O(n)$ for the adjacency list, $O(n)$ for the $size$, $down$, and $answer$ arrays. Total space is $O(n)$.

Compare this to the brute force $O(n^2)$ approach. For $n = 200{,}000$, rerooting finishes in milliseconds while brute force would time out.