Why is baryon or lepton violation in standard model is a non-perturbative effect? The baryon number B or lepton number L violation in the standard model arise from triangle anomaly. Right? Triangle diagrams are perturbative diagrams. Then why the B or L violation in Standard model is said to be a non-perturbative effect? I'm confused.
 A: It is a non-perturbative effect because it is 1-loop exact. 
The triangle diagram is actually the least insightful method to think about this, in my opinion. The core of the matter is the anomaly of the chiral symmetry, which you can also, for example, calculate by the Fujikawa method examining the change of the path integral measure under the chiral transformation. You can obtain quite directly that the anomaly is proportional to
$$\int \mathrm{Tr} (F \wedge F)$$
which is manifestly a global, topological term, (modulo some intricacies) it is the so-called second Chern class and takes only values of $8\pi^2k$ for integer $k$. It is, by the Atiyah-Singer index theorem (this can also be seen by Fujikawa), essentially the difference between positive and negative chiral zero modes of the Dirac operator. This is obviously a discontinuous function of $A$ (or $F$), which is already bad for something which, if it were perturbative, should be a smooth correction to something, and it is also the number describing which instanton vacuum sector we are in, see my answer here. Since perturbation theory takes place around a fixed vacuum, this is not a perturbative effect, since it is effectively describing a tunneling between two different vacuum sectors.
