Your rule should be amended to require that $t_1$ and $t_2$ are _disjoint_ binary trees, i.e., they have no nodes in common. Then I trust the answer to your question becomes obvious.
Your rule should be amended to require that $t_1$ and $t_2$ are _disjoint_ binary trees, i.e., they have no nodes in common. Then I trust the answer to your question becomes obvious.