Why is this an NFA, but not an FA?

Question

It doesn't have non-deterministic paths.

Every transition is one-to-one.

Accepted states are defined/guaranteed on specific inputs.

Not sure why its not an FA

Edit: 0.0.2

Answer 1

Nvm, figured it out.

The transition function $\delta$ must be total in a FA.

In the 1st example, at the accepting state $q_1$, $\delta(q_1, 0)=undefined$.

Whereas in the 2nd example, if we label the states $q_0, q_1,q_2$ from left to right, at the accepting state $q_1$, $\delta(q_1, 0)=q_2$.