It's not clear for me how to represent the proof tree of a sequent that doesn't use and hypothesis, for example: $p \vdash q \rightarrow p$. The problem is that $q$ should appear as hypothesis to derive $q \rightarrow p$, eventhough is not used.
First try (bad bad bad try), clearly the premise $p$ should be a leaf, the tree makes it look as if it's derived from $[q]$: tree proof 1.
Second try (better or worse, who knows?), the hypothesis is left as a dangling node: tree proof 2.
$q$ should appear as hypothesis to derive $q→p$, eventhough is not used.
No, it shouldn't. With rules that allow to discharge assumptions, there is no requirement that this assumption actually be discharged, or even occur anywhwere in the derivation. $\dfrac{p}{q \to p}$ is a legal use of the $\to i$ rule, and all you need to prove your sequent.
As has been pointed out, so-called vacuous discharge is standardly allowed. Even if $q$ is never used in a proof that ends in $p$, we are allowed to invoke Conditional Proof and go on to derive $q \to p$.
But, if it bugs you, you could use $q$, as in the first inference step here:
$\quad\quad p\quad\quad\quad [q]\\ \quad\quad\quad p \land q\quad\quad\quad \land I\\ \quad\quad\quad\ \ \ p\quad\quad\quad\ \ \ \land E\\\\ \quad\quad\quad q \to p\quad\quad\quad CP$
But that just goes to show that if you were to try to ban vacuous discharge, you'd have also ban some other proof steps combining to have the same effect.
(Which can be done: but then you'd need to think through why one might want to ban vacuous discharge given that it is evidently classically (and intuitionistically) valid.)
p[q]^1}{q\to p~~}{\small({\to}\mathcal I^1)}$$ – Graham Kemp Sep 01 '23 at 00:29