I tried it as follows and would like to know if it is correct.

I tried it as follows and would like to know if it is correct.

The analysis is not accurate although the result is correct. You could write it more accurately by replacing $=$ with $\le$
$T(n) \le c(1+2+..+2^{n-1})$ ( $\le$ since not all level have same number of children, consider the most right-handed path, n is decreasing by $2$ every step ).
Indeed a more careful analysis can get you a tighter bound as mentioned in the comment. The idea is, the time $T(n)$ is computed with $T(n-1) + T(n-2)$ the same way as the actual fibonacci $F(n)$, and since $F(n) = O(\phi^n)$ for $\phi = (1+\sqrt{5})/2$ as the closed form.
Thus $T(n) = O(\phi^n)$ which is slightly smaller than $2^n$