Relation between digraph and NP-Complete problem
When a problem is NP-complete, it doesn't necessarily mean that all instances are hard to solve. It just means that enough instances are hard to solve that, if you could solve this problem efficiently, you could do the same for all problems in NP. Enough, in this case, does mean infinitely many1 and that should lead you towards looking at large inputs if you want to find hard cases.2

For example, the fact that 3-SAT is NP-complete doesn't mean that you would have any difficulty telling me whether the formula $(X\vee Y\vee Z)$ is satisfiable. In your example, you've observed that, for a graph with a single edge, that edge is a Hamiltonian path. That doesn't mean that it's hard to tell whether a graph has a single edge, or that it's easy to compute Hamiltonian paths on all graphs.



1 Probably infinitely many, anyway. If P&nbsp;=&nbsp;NP, there wouldn't be any "hard" instances of NP-complete problems.

2 Note that, strictly speaking, there's no such thing as a single "hard case" since, if we restrict to that single input, the problem can be solved in constant time. Properly, I mean a whole infinite class of instances such that it's "hard" to write an algorithm that answers the problem on that whole class of inputs.