Intro

You know how to run DFS from a fixed root and compute subtree answers. But what if the problem asks you to compute that answer for every possible root? Running a separate DFS from each node gives $O(n^2)$, which is too slow for $n = 200{,}000$.

The rerooting technique solves this in $O(n)$. You compute the answer for one root, then shift it to every other root by adjusting contributions as you move along edges. I'll walk you through the two-pass algorithm, show you the template, and apply it to problems from CSES and LeetCode.