  Let $R$ and $B$ be two disjoint sets of points in the plane such that
$|B|\leqslant |R|$, and no three points of $R\cup B$ are collinear. We show
that the geometric complete bipartite graph $K(R,B)$ contains a non-crossing
spanning tree whose maximum degree is at most $\max\left\{3, \left\lceil
\frac{|R|-1}{|B|}\right\rceil + 1\right\}$; this is the best possible upper
bound on the maximum degree. This solves an open problem posed by Abellanas et
al. at the Graph Drawing Symposium, 1996.
