Down Pass (Subtree distances)

Pick node $1$ as root. For each node $v$, compute down[v] = sum of distances from $v$ to all nodes in its subtree. Also track size[v] = number of nodes in subtree of $v$. Recurrence: down[v] = Σ_(u in children(v)) (down[u] + size[u]). The down[u] part is the sum within $u$'s subtree. The +size[u] part accounts for the extra edge from $v$ to $u$ for each of the size[u] nodes. After this pass, down[1] equals the answer when node $1$ is root.

But you need answers for all other roots. That requires the up pass.