It is well known that : $$\binom{n}{0} + \binom{n}{2} + \cdots = \binom{n}{1} + \binom{n}{3} + \cdots = 2^{n-1}$$

Therefore, we have $$\begin{align*} &\binom{n}{0} - \frac{1}{2}\binom{n}{1} + \binom{n}{2} - \frac{1}{2}\binom{n}{3} + \cdots \\ =\ & \left(\binom{n}{0} + \binom{n}{2} + \cdots\right) - \frac{1}{2}\left(\binom{n}{1} + \binom{n}{3} + \cdots\right) \\ =\ &2^{n-1}-\frac{1}{2}\times 2^{n-1} \\ =\ & \boxed{2^{n-2}} \end{align*}$$