For a labeled tree $T$ on $\{1,2,...,n\}$ an *inversion* of $T$ is a pair $1 < i < j \leq n$ such that $j$ belongs to the unique path from $1$ to $i$ (we think of $T$ as being rooted at $1$). Let $\mathrm{inv}(T)$ denote the number of inversions of $T$.


Define the generating function $f(q) := \sum\_{T} q^{\mathrm{inv}(T)}$ where the sum is over all labeled trees on $\{1,2,...,n\}$.


Then it is known that $f(-1)$ is the number of [alternating permutations](https://en.wikipedia.org/wiki/Alternating_permutation) in $\mathfrak{S}\_n$ (i.e., the so-called "Euler number"). See e.g. Goulden-Jackson 3.3.49(d).


**Question**: Is there a simple proof of this result via a sign-reversing involution?



Sign-reversing involution proof of tree inversion generating function at $-1$ equals number of alternating permutations?
J.-J. Pansiot, [Nombres d'Euler et Inversions dans les Arbres](https://www.sciencedirect.com/science/article/pii/S0195669882800371), Europ. J. Combin. 3 (1982), 259–262, uses a sign-reversing involution to show that $f(-1)$ is the number of increasing trees on $[n]$ in which every vertex other than the root has an even number of children. In Pansiot's paper he gives a reference to a paper of Viennot (that I have not looked at) that shows that these trees are counted by the Euler numbers. 


