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My first solution worked but was way too slow, and they pushed me to optimize it.
Use a two-pass DFS technique: first compute the answer for an arbitrary root (e.g., node 0) via post-order traversal, then perform a pre-order traversal to 'reroot' the tree, updating the answer for each child based on the parent's answer. This achieves O(n) time by avoiding recomputation for each root.
Pro tip: Emphasize that the rerooting technique works for any associative and commutative aggregation (like sums, counts, max/min) and that the key is to define a transition function that updates the answer when moving the root from a parent to a child.
Clarify what 'answer' means for a given root (e.g., sum of distances to all nodes, subtree sizes, etc.). Identify if the answer can be computed from children's answers and the node itself.
Pick any node as root (e.g., 0) and perform a post-order DFS to compute the answer for that root. Also compute any auxiliary data needed for rerooting (e.g., subtree sizes, sums).
Determine how the answer changes when the root moves from a parent to a child. For example, for sum of distances, the child's answer = parent's answer + (n - 2*subtree_size[child]).
Traverse the tree from the initial root, and for each child, compute its answer using the transition from the parent's answer. Store the result for each node.
Confirm O(n) time and O(n) space. Discuss handling of large trees, recursion depth (use iterative DFS if needed), and special cases like n=1.
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