Question: We can use DFS on an undirected graph by considering it directed by making each undirected edge as two anti-parallel directed edges. As we have

We can use DFS on an undirected graph by considering it directed by making each undirected edge as two anti-parallel directed edges. As we have seen, DFS produces a rooted tree. Given a rooted tree T subgraph of G, could it have been produced by a DFS of G starting at the root.

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