Let T be a depth-first search tree in an undirected graph G. Vertices u and ν are leaves of this tree T. The degrees of both u and ν in G are at least 2.

Let's interpret question correctly and draw inferences

**Vertices u and ν are leaves of this tree T.**

Means (1) (**u,v) is not an edge in the graph **otherwise one would be have been descendent of another and both of them must not be leaves.

(2)If vertices u and v are leaves of tree T, then when DFS was exploring them, after exploring say vertex u, it was unable to find any new unvisited neighbour of u and hence, it had to backtrack the search.So u became leave of T. Same story goes with vertex v.

(3)If degrees of vertices u and v are at least 2 then I say consider the scenario for vertex u

Consider some intermediate vertices K and L which are neighbours of vertex U, and there may be more vertices than K and L, ahead of them connected to either one of them(to K or L) but for simplicity, I consider only two(K and L).

Now, say my DFS algorithm explored vertex K, coloured it, Grey, then it went to vertex U and found it WHITE (means unvisited) so it marks it Grey and makes the edge (K,U) a tree edge. Now DFS examines neighbour of U and finds neighbour L.

Now if the neighbour L was WHITE (Means not visited), DFS would have visited this vertex L and then edge (U,L) would have been marked as tree edge and in this case Vertex, U would no longer be a leave in DFS Tree T.

So, Necessarily my vertex L was visited before U (L Maybe GREY or BLACK in colour) and this would have been connected to vertex K via some other vertices or directly, otherwise, it was impossible for this vertex L to be discovered via path any other than from U to L.

So, what all this means?

Yes Vertex U and it's neighbour are in a cycle and your DFS will mark some edges as back edges.

The similar story would hold for vertex V.

So, even if Option had said

"Vertex U and V are involved in cycle with their neighbour" then also this would have been true.

**So Option (D) is the answer.**