Here is some enrichment material to complete this calculation. First
note that these planted plane trees correspond to ordinary plane trees
with an extra node attached at the root.
The species equation for these enumerating by the internal nodes
(i.e. excluding the node where the tree is planted) is
$$\mathcal{T} = \mathcal{Z}\times \mathfrak{S}(\mathcal{T}).$$
This gives the functional equation
$$T(z) = z\frac{1}{1-T(z)}
\quad\text{or}\quad
z = T(z)(1-T(z)).$$
Now to extract coefficients from this by Lagrange inversion use
$$[z^n] T(z)
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} T(z)\;dz$$
and put $w=T(z)$ where $dz = 1-2w\; dw$ to obtain
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{n+1} (1-w)^{n+1}}
w \times (1-2w) \;dw$$
which is
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\left(\frac{1}{w^n (1-w)^{n+1}}
- 2\frac{1}{w^{n-1} (1-w)^{n+1}} \right) \; dw.$$
This yields
$${n-1+n\choose n} - 2{n-2+n\choose n}
= {2n-1\choose n} - 2{2n-2\choose n}$$
which is
$$\frac{2n-1}{n}{2n-2\choose n-1}
-2\frac{n-1}{n}{2n-2\choose n-1}
= \frac{1}{n} {2n-2\choose n-1}.$$
These are of course the Catalan numbers.
The species equation for these trees with the even outdegree marked is
$$\mathcal{Q} =
\mathcal{Z}\times
\mathcal{U}\mathfrak{S}_\mathrm{even}(\mathcal{Q})
+ \mathcal{Z}\times
\mathfrak{S}_\mathrm{odd}(\mathcal{Q}).$$
This gives the functional equation
$$Q(z) = uz\frac{1}{1-Q(z)^2}
+ z\frac{Q(z)}{1-Q(z)^2}$$
or
$$Q(z)(1-Q(z)^2) = uz + z Q(z).$$
To compute the total number of even degree nodes introduce
$$G(z) = \left.\frac{\partial}{\partial u} Q(z)\right|_{u=1}.$$
Differentiate the functional equation and put $u=1$ to get
$$G(z)(1-T(z)^2) + T(z) (-2T(z) G(z))
= z + z G(z).$$
This yields
$$G(z) = \frac{z}{1-z-3T(z)^2}.$$
To extract coefficients from $G(z)$ use
$$[z^n] G(z)
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\frac{z}{1-z-3T(z)^2}\;dz.$$
Using the same substitution as before we obtain
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{n+1} (1-w)^{n+1}}
\frac{w(1-w)}{1-w(1-w)-3w^2} \times (1-2w) \;dw$$
which is
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^{n+1} (1-w)^{n+1}}
\frac{w(1-w)}{(1-2w)(1+w)} \times (1-2w) \;dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^n (1-w)^n}
\frac{1}{1+w}\;dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{1}{w^n (1-w)^n}
\sum_{q\ge 0} (-1)^q w^q \; dw.$$
Extracting coefficients from this we obtain
$$\sum_{q=0}^{n-1} {q+n-1\choose n-1} (-1)^{n-1-q}.$$
This gives OEIS A026641 which is
$$1, 1, 4, 13, 46, 166, 610, 2269, 8518, 32206,\ldots$$
where we find the above workings confirmed.
It follows that the average number of even outdegree
nodes in a random rooted plane tree is given by the formula
$$ n \times {2n-2\choose n-1}^{-1} \times
\sum_{q=0}^{n-1} {q+n-1\choose n-1} (-1)^{n-1-q}.$$