  We say a finite poset $P$ is a tree poset if its Hasse diagram is a tree. Let
$k$ be the length of the largest chain contained in $P$. We show that when $P$
is a fixed tree poset, the number of $P$-free set systems in $2^{[n]}$ is
$2^{(1+o(1))(k-1){n \choose \lfloor n/2\rfloor}}$. The proof uses a
generalization of a theorem by Boris Bukh together with a variation of the
multiphase graph container algorithm.
