Wow, it's more straightforward than I thought. Thanks a lot!

Thanks for your brilliant answer every time.

Now I fully understood the proof. Thanks a lot!

It may seem obvious, but can I ask one more thing? Here we fixed $\ell$ and found that $n=6\ell-2$ is the smallest $n$ with the given condition. But can we say $\ell=\frac{n+2}{6}$ is the largest $\ell$ with fixed $n$?

Thanks a lot! But I have one more question: If $n \equiv 2$ (mod $6$) or $n \equiv 0$ (mod $6$), there are $4$ or $2$ remaining vertices. We should replace one gadget to bigger block, or add them to $T$. I think the former is reasonable way, since the latter makes $T$ no more tree. Then how can we assure that $G$ is the smallest?

@PeterTaylor Thanks. Now I revised it.