Suppose $X \sim Bin(n,p)$ and $Y \sim Bin(n,1-p)$, how is $X+Y$ distributed?

Assume a tree with binary offspring. Assume the root is generation 0. Take last generation $X_{n,1},\ldots,X_{n,2^n}$ be the bernoulli variables for generation, let $N_n = \sum X_{n,i}$, let $\phi_p$ be the characteristic function for $Bin(2,p)$, and $\phi_q$ the moment generation function for $Bin(2,q)$. Then we have (as can be seen by conditioning on the last generation) $$ E(\exp(sN_n)) = E\phi_p(s)^{X_{n-1,1}}\phi_q(s)^{1-X_{n-1,1}}\cdots = E\left(\phi_q(s)^{2^{n-1}}\left ( \frac{\phi_p(s)}{\phi_q(s)} \right ) ^{\sum_i X_{n-1,i}}\right) $$ Now this can be rewriten as $$ CE\exp(s_{n-1}N_{n-1}) $$ with $$ C = \phi_q(s)^{2^{n-1}} $$ and $$ s_{n-1} = \ln \left (\frac{\phi_p(s)}{\phi_q(s)} \right ) $$ Now use induction. If one is after probabilities consider using https://en.wikipedia.org/wiki/Probability-generating_function and note $$ Ez^N = E(\exp(ln(z) N)) $$