There is a bijection ${\mathcal{T}}_n\to {\mathcal{D}}_n$ for every $n$

[base] The set ${\mathcal{T}}_0$ has the single tree $\{\epsilon \}$ and ${\mathcal{D}}_0$ has the single empty path, so ${\mathrm{\Phi }}_0$ sending $\{\epsilon \}$ to the empty path is a bijection.

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

For a tree $T\in {\mathcal{T}}_{n+1}$ write its recursive data as $(i,L,R)$ as in [F1], with $L\in {\mathcal{T}}_i$ and $R\in {\mathcal{T}}_{n-i}$, and define ${\mathrm{\Phi }}_{n+1}(T)$ to be the Dyck path whose step word is $U$, then the step word of ${\mathrm{\Phi }}_i(L)$, then $D$, then the step word of ${\mathrm{\Phi }}_{n-i}(R)$. This lands in ${\mathcal{D}}_{n+1}$ by the defining condition on Dyck paths.

For a Dyck path $Q\in {\mathcal{D}}_{n+1}$, the first-return factorisation of [L1] writes $Q$ uniquely as $UPD{Q}^{\mathrm{\prime }}$ with $P\in {\mathcal{D}}_i$ and $Q^{\mathrm{\prime }}\in {\mathcal{D}}_{n-i}$ for a unique $i\le n$, so the induction hypothesis supplies unique trees $L:={\mathrm{\Phi }}_i^{-1}(P)$ and $R:={\mathrm{\Phi }}_{n-i}^{-1}(Q^{\mathrm{\prime }})$ and therefore a unique tree $T$ with recursive data $(i,L,R)$. Define ${\mathrm{\Psi }}_{n+1}(Q):=T$.

The definitions of ${\mathrm{\Phi }}_{n+1}$ and ${\mathrm{\Psi }}_{n+1}$ undo one another: starting from a tree, the factorisation recovered from its image is the same recursive split, and starting from a Dyck path, the tree recovered from its first return rebuilds the same path. Hence ${\mathrm{\Psi }}_{n+1}\circ {\mathrm{\Phi }}_{n+1}={\mathrm{\Delta }}_{{\mathcal{T}}_{n+1}}$ and ${\mathrm{\Phi }}_{n+1}\circ {\mathrm{\Psi }}_{n+1}={\mathrm{\Delta }}_{{\mathcal{D}}_{n+1}}$, so ${\mathrm{\Phi }}_{n+1}$ is a bijection by [L2].