# Sign-reversing involution proof of tree inversion generating function at $-1$ equals number of alternating permutations?

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 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?

J.-J. Pansiot, Nombres d'Euler et Inversions dans les Arbres, 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.