The question is as follows: True or False: For every non-directed connected non-weighted graph and for every spanning tree T of the graph there exists a vertex v such that T is a DFS tree with the root v.
What about if instead of DFS I used BFS?
I have no clue where to begin with this one. I feel like I'm overlooking some basic characteristic of the algorithm or the tree that it produces. Any help would be appreciated!