Suppose G is a binary tree. Is G necessarily planar? Give a proof, or a counterexample.
My guess is that it is indeed planar but I am struggling to find a formal proof for this.
EDIT: Is there a proof that does not use Kuratowski's theorem? It was an exam question and we are not supposed to know that.
Indeed it is true by Kuratowski's theorem or by the following argument which I will sketch below.
We induct on the number of vertices of the tree, $G$, the case where $G$ has only one vertex being obvious. Every tree has a vertex with degree one, call it $v$ and call the vertex it is connected to $u$. Remove $v$ and by induction, $G - v$ can be drawn in the plane with no crossings. Now in a small enough neighborhood surrounding $u$ one will only find the edges that contain $u$ as an endpoint. In this neighborhood it is easy to put the edge $(u,v)$ in without creating any crossings.