Here's a semi-explicit construction for a smooth function f that is zero precisely on the classical Cantor set. By this set I mean the one that is obtained from I_0 = [0,1] by repeatedly removing the middle third of any ensuing interval. So let's denote by I_n the
n-th set in this process.

Now let's make a smooth function f_n on [0,1] such that its zero set is exactly I_n.
Starting with f_0 = 0 we obtain f_{n+1} from f_n as follows:

set f_{n+1} = f_n on I_{n+1} and
on an interval that is removed from I_n make f_{n+1} equal to a bump function
that is 0 only at the boundary of the interval. We can choose the bump function to be of height 2^{-2^n}.

This choice of heights of the bump functions will ensure that
the derivatives of f all converge uniformly to their pointwise limits.
Hence the limit function f_n is again smooth. By construction its zero set is exactly
the Cantor set.