False. Take edges and , and let the outer DFS loop try vertices in order . The first call finishes immediately. The next discovers , whose only outgoing edge reaches already black , so its tree contains only . Finally sees already black .
Vertex has both an incoming edge from and an outgoing edge to , yet is the sole vertex of its DFS tree. Nonzero incoming and outgoing degrees do not require either edge to become a tree edge in depth-first search.
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