There are at least two different definitions of Dyck path; I prefer to think of them as mountain ranges, i.e., paths from $\langle 0,0\rangle$ to $\langle 2n,0\rangle$, using steps $\langle 1,1\rangle$ and $\langle 1,-1\rangle$, that do not drop below the $x$-axis. If you think of them as lattice paths from $\langle 0,0\rangle$ to $\langle n,n\rangle$ that do not rise above the diagonal line $y=x$, it’s not hard to make the conversion: your step to the right is my up-step, your step up is my down-step, and the diagonal line corresponds to my $x$-axis.
Let $P$ be a Dyck path of length $2(n+1)$, and let $\langle 2k,0\rangle$ be the first point to the right of $\langle 0,0\rangle$ at which $P$ hits the $x$-axis. The part of $P$ from $\langle 0,0\rangle$ to $\langle 2k,0\rangle$ is a Dyck path of length $2(k-1)$ preceded by an up-step and followed by a down-step, and the part of $P$ from $\langle 2k,0\rangle$ to $\langle 2(n+1),0\rangle$ is a Dyck path of length $2(n+1-k)$. There are $C_{k-1}$ Dyck paths of length $2(k-1)$, and there are $C_{n+1-k}$ Dyck paths of length $2(n+1-k)$. Finally, $k$ ranges from $1$ through $n+1$, so
$$C_{n+1}=\sum_{k=1}^{n+1}C_{k-1}C_{n+1-k}=\sum_{k=0}^nC_kC_{n-k}\;.$$
A picture will help to visualize the bijection described in this post : please take your pencil.
First, the setting : it will be easier to view Dyck paths as (ordered rooted) trees, using the traditional device of the contour function (I put a ref below). A peak is then a leaf of the tree (a vertex without children). Also, Dyck paths being of length 2𝑛, the trees will have n edges, so that the minimal number of leaves will be 1 and the maximal number will be 𝑛. Call brothers of a vertex the children of the same father (by convention, we discard the vertex itself).
Your claim is that, among trees on $n$ edges, there is as many with $k$ leaves as with $n+1-k$ leaves.
The bijection goes as follows : to a (rooted) tree on $n$ edges, we will associate another (rooted) tree with $n$ edges, by keeping the same vertex set (also the same root) and changing the edge set. To that end, it is enough to define the father of every vertex distinct from the root in the new tree, let's call it the new father.
Consider a vertex distinct from the root and look for the most recent ancestor of that vertex that has a brother to its right : this will the vertex itself if it has a brother to its right, or the father of this vertex if the vertex itself has no brother to its right but its father has, and so on...
The converse map is obtained by replacing the word "right" by the word "left" in the above description.
I'll let you check this works with respect to the property you have in mind for the leaves : a possible proof is by recurrence on the number of leaves (it is easy to see it works for one leave, then one looks at the shape of trees with two leaves, and from there, the idea os the recurrence should be quite clear)
(This is closely related, but in a subtle way distinct from what is called rotation bijection in computer science, see Flajolet and Sedgewick on p.73.)
--
The reference on the contour function (in French sorry):
https://fr.wikipedia.org/wiki/Arbre_de_Galton-Watson#Processus_de_contour.
--
From far, the question of MathSE on which I spent the most time. Perhaps it is known in combinatorics, I have done no research on the topic.
Best Answer
I find it a bit easier to think in terms of paths from $\langle 0,0\rangle$ to $\langle 2n,0\rangle$ that consist of $n$ up-steps (steps from $\langle k,\ell\rangle$ to $\langle k+1,\ell+1\rangle$) and $n$ down-steps (steps from $\langle k,\ell\rangle$ to $\langle k+1,\ell-1\rangle$). An up-step in this version corresponds to a step to the right in your version, and a down-step corresponds to a step upwards in your version. Your boundary condition becomes a requirement that my path not drop below the line $y=-z$.
We can use a slight modification of one of the usual arguments for counting the paths that don’t drop below the line $y=0$.
As in your version, there are altogether $\binom{2n}n$ paths from $\langle 0,0\rangle$ to $\langle 2n,0\rangle$, and the problem is to count the ‘bad’ ones, i.e., the ones that do drop below the line $y=-z$. Suppose that we have a bad path $\pi$. There is a first point at which $\pi$ reaches the line $y=-z-1$; if it has made $u$ up-steps at that point, it must have made $u+z+1$ down-steps and so have reached the point $\langle 2u+z+1,-z-1\rangle$. Reflect the remainder of $\pi$ (i.e., the part to the right of this point) in the line $y=-z-1$. That part of $\pi$ has $n-u$ up-steps and $n-u-z-1$ down-steps, so its reflection has $n-u$ down-steps and $n-u-z-1$ up-steps. This means that it must end at the point
$$\langle 2u+z+1,-z-1\rangle+\langle2n-2u-z-1,-z-1\rangle=\langle 2n,-2z-2\rangle\;.$$
Conversely, any path from $\langle 0,0\rangle$ to $\langle 2n,-2z-2\rangle$ must hit the line $y=-z-1$, and if we reflect the part of it to the right of that intersection in the line $y=-z-1$, we get a path from $\langle 0,0\rangle$ to $\langle 2n,0\rangle$ that drops below the line $y=-z$. Thus, we have a bijection between our bad paths and all paths from $\langle 0,0\rangle$ to $\langle 2n,-2z-2\rangle$. Each of these paths has $n-z-1$ up-steps and $n+z+1$ down-steps, so there are $\binom{2n}{n+z+1}$ of them. Thus, there are
$$\binom{2n}n-\binom{2n}{n+z+1}=\binom{2n}n-\binom{2n}{n-z-1}$$
good paths from $\langle 0,0\rangle$ to $\langle 2n,0\rangle$.