 Solution

There are exactly two tree Feynman diagrams: the incident scalar can be absorbed before the final scalar is emitted, or after it. The internal fermion four-momenta are respectively $r_s=p+k=q+l$ and $r_u=p-l=q-k$. A scalar exchange diagram would require an absent scalar self-interaction, so there is no additional tree channel.

Define the S-matrix convention $S_{fi}={\delta }_{fi}+i(2\pi {)}^4{\delta }^{(4)}(p+k-q-l)T_{s^{\prime }s}$. The two tree scattering amplitudes areUsing the Dirac propagator from the preceding Feynman rules,The two terms add with the same relative sign: neither diagram exchanges identical external fermions. Their interference must be retained when squaring the complete scattering amplitude. In the phase convention above, ${\gamma }^5(M-i/r){\gamma }^5=-(M+i/r)$, which gives an equivalent simplified numerator. An overall phase depends on the S-matrix convention and does not change a relativistic scattering cross-section.