Let $ G $ be a tree and $ V_0 = \left \lbrace v \in V(G) | deg (v) = 1 \right \rbrace $. Show that $ G \setminus V_0 $ is a tree.
To see that this is true, to demonstrate:
$ G \setminus V_0 $ is acyclic. Suppose $ G \setminus V_0 $ contains a cycle $ C = (u_1, \: \ldots, \: u_n) $, but $ G \setminus V_0 $ is a subplot of $ G $, then $ C \subset G $ , which is a contradiction. Therefore $ G \setminus V_0 $ is acyclic.
$ G \setminus V_0 $ is connected. Let $ u, \: v \in V (G \setminus V_0) $, since $ G $ is connected then there is $ T $ a $ uv-$path in $ G $. Let $ T = (u = x_0, \: \ldots, \: x_k = v) $, but $ deg (x_i) \geq 2 $ for all $ i = 1, \: \ldots, \: k-1 $ so $ T $ is contained in $ G \setminus V_0 $ so there is a $ uv-$path. For this, we must use what we know, $ G $ is connected, but I still have doubts...