Dear liu,
If $A$ is a domain in which every finitely generated ideal is principal, then a module over $A$ is flat iff it is torsion free (Bourbaki, Comm.Alg.,I,§2, 4, Prop.3). Of course a PID has this property, but the ring $\mathcal O(U)$ of holomorphic functions over a connected open subset $U\subset \mathbb C$ also has it, although it is not a PID.[Ah, I have just checked the definition of Prüfer and it seems that these rings are actually Prüfer, so you knew this. I'll leave this class of examples for the benefit of those who, like me, don't know the concept "Prüfer ring".]
A simple example of torsion free non flat module $E$ over a ring $A$ is $A=\mathbb C [t^2, t^3] \subset E=\mathbb C [t]$. This corresponds to the normalization of the cusp $S=Spec (\mathbb C[X,Y]/(Y^2-X^3))$ i.e. to the morphism $f: \mathbb A ^1 \to S \subset \mathbb A ^2$ given by $x=t^2, y=t^3$. Non-flatness is due to the fact that the fiber of $f$ at the origin is a double point on $\mathbb A ^1$ (supported at the origin) whereas the other fibers are simple points. This is a tiny particular case of the constancy of Hilbert polynomials in flat families. Of course there are direct proofs by pure algebra, but I hope you like geometry...