We have schedule S with transactions $T_1,T_2.....T_n$
Question says
If S is serializable
Now S can be either Conflict serializable or Can be view Serializable but Not conflict serializable.
Take an example of below schedule
T1 T2
W(A)
W(A)
W(A)
The precedence graph of above will be cyclic and the above schedule is not Conflict serializable , but View serializable as T2->T1 and hence SERIALIZABLE.
So, clearly my topological sort algorithm would fail on such a precedence graph.
DFS algorithm would only tell about the nature of precedence graph( Like whether it has cycle or not, ancestor-descendants relationship) and this has nothing to do with serializability.Similarly for BFS.
Option (D) is clearly non-sense
Now if I am able to produce a Topological order of a precedence graph, that means the graph is Acyclic and hence the schedule S will be conflict equal to the serial schedule order given by topological order.A conflict equivalent schedule is also view equivalent and hence serializable.
Answer (A).