TITLE: Collecting alternative proofs for the oddity of Catalan
QUESTION [15 upvotes]: Consider the ubiquitous Catalan numbers $C_n=\frac1{n+1}\binom{2n}n$. In this post, I am looking for your help in my attempt to collect alternative proofs of the following fact: $C_n$ is odd if and only if $n=2^r-1$.
Proofs that I know:

an application of Legendre's (Kummer) formula $\nu_2(C_n)=s(n+1)-1$, where $s(n)$ is the sum of the binary digits of $n$.

See E. Deutsch and B. Sagan, Congruences for Catalan and Motzkin numbers and related sequences.

See A. Postnikov and B. Sagan, What power of two divides a weighted Catalan number?

See O. Egecioglu, The parity of the Catalan number via lattice paths.



QUESTION. Do you know of or can provide a new proof that $C_n$ is odd if and only if $n=2^r-1$?

REPLY [37 votes]: Taking the fact that Catalan numbers $C_n$ measure the number of binary trees on $n$ nodes, we can find an involution on the set of these trees: choose the lexicographically first node in the tree that's unbalanced (i.e., where the number of nodes in the left subtree is different from the number in the right subtree). Then partner this tree with the one where we swap the left and right subtrees. This involution is fixed-point-free except in the case where no node in our tree is unbalanced — that is, when we have the complete binary tree of $n=2^t-1$ nodes, where we have to pair this tree to itself.