There is a bijection ${\mathcal{T}}_n\to {\mathcal{P}}_{n+2}$ for every $n\in \mathbb{N}$

[base] The set ${\mathcal{T}}_0$ has the single tree $\{\epsilon \}$ and the set ${\mathcal{P}}_2$ has the single empty triangulation, so there is a unique bijection ${\mathrm{\Psi }}_0:{\mathcal{T}}_0\to {\mathcal{P}}_2$.

[ih] Assume that for every index $j\le n$ a bijection ${\mathrm{\Psi }}_j:{\mathcal{T}}_j\to {\mathcal{P}}_{j+2}$ has already been constructed.

For a tree $T\in {\mathcal{T}}_{n+1}$ write its recursive data as $(i,L,R)$ as in [F1]. Let ${\mathrm{\Psi }}_{n+1}(T)$ be the triangulation of the $(n+3)$-gon obtained by taking the triangle on the closing side with third vertex $k:=i+2$, filling the left $k$-gon by ${\mathrm{\Psi }}_i(L)$, and filling the right $(n-i+2)$-gon by the order-preserving relabelling of ${\mathrm{\Psi }}_{n-i}(R)$ onto the vertices $\{k,k+1,\dots ,n+3\}$.

For a triangulation $U\in {\mathcal{P}}_{n+3}$, [L1] supplies a unique split index $k$ and therefore a unique index $i:=k-2\le n$, together with triangulations of the left $k$-gon and the right $(n-i+2)$-gon. Relabel those two sub-polygons back to $\{1,\dots ,i+2\}$ and $\{1,\dots ,n-i+2\}$, apply the inverse bijections ${\mathrm{\Psi }}_i^{-1}$ and ${\mathrm{\Psi }}_{n-i}^{-1}$ from the induction hypothesis, and rebuild a tree in ${\mathcal{T}}_{n+1}$ from the recursive data $(i,L,R)$. Define that tree to be ${\mathrm{\Omega }}_{n+1}(U)$.

The constructions in steps 2.1 and 2.2 undo one another because both are governed by the same split index: the root split of the tree becomes the closing-side triangle of the triangulation, and the closing-side triangle of the triangulation becomes the root split of the tree. Hence ${\mathrm{\Omega }}_{n+1}\circ {\mathrm{\Psi }}_{n+1}={\mathrm{\Delta }}_{{\mathcal{T}}_{n+1}}$ and ${\mathrm{\Psi }}_{n+1}\circ {\mathrm{\Omega }}_{n+1}={\mathrm{\Delta }}_{{\mathcal{P}}_{n+3}}$, so ${\mathrm{\Psi }}_{n+1}$ is a bijection by [L2].