On the one hand, the comments explain that a low-growth-rate tree
can still have continuum many branches. Indeed, the growth rate
can be extremely slow, with most levels having no splitting at
all, but then every once in a very long while, a single node
splits. Just make sure at the $k^{th}$ such splitting that you are
splitting at a node above the $k^{th}$ binary sequence of the tree
(in some fixed standard enumeration of all finite binary
sequences), and then the resulting tree $T$ will have the property
that every node lies below a splitting node. This property will
ensure a subtree of type $2^{<\omega}$, and hence give you
continuum many branches.

In contrast, it is the converse question that I find extremely
interesting. Namely,

**Question.** What is a sufficient growth rate on the tree $T$ to
ensure that it has continuum many branches?

The answer is that the growth rate must be essentially close to
$2^n$, in the precise senses described by the following theorems.
On the one hand, such a high growth rate suffices for continuum
many paths:

**Theorem.** If $T\subset 2^{<\omega}$ is a binary tree with
growth rate $\Omega(2^n)$ (in the Knuth sense [Big-Omega notation](http://en.wikipedia.org/wiki/Big_O_notation#Related_asymptotic_notations)), then $T$
has continuum many branches.

Proof. What I intend to assume on the growth rate is that is that
there is a positive real number $r>0$ such that the number of
nodes on the $n^{th}$ level of $T$ is at least $r2^n$. Thus, the
complement of the set of paths through $T$ is approximated by the
unions of the cones above the omitted nodes on this level, and
this has measure at most $1-r$ with respect to the usual
coin-flipping probability measure on Cantor space. Since this is
true at each level, it follows that the measure of the paths
through $T$ is at least $r$, which is positive, and so there must
be uncountably many paths through $T$. Since this is a closed set,
it follows by the [Cantor-Bendixson theorem](http://en.wikipedia.org/wiki/Perfect_set_property) that $T$ has
continuum many paths. QED

On the other hand, any lower growth rate does not ensure continuum many paths.

**Theorem.** If $f:\mathbb{N}\to\mathbb{N}$ is $o(2^n)$ (see
[Little-o notation](http://en.wikipedia.org/wiki/Big_O_notation#Little-o_notation)),
then there is a tree $T$ with growth rate
exceeding $f$, but having only countably many paths.

Proof. We build $T$ in stages. First, we let $T$ use the full
binary branching up to some level $n_0$. Then, at this stage, we
allow only half of the nodes to branch, the ones that started out
with $1$ in the their first bit, and force the other half, the
ones beginning with $0$, to become isolated branches in the tree
starting at level $n_0$, continued only with more $0$s after
$n_0$. Meanwhile, we let the other "live" nodes, beginning with
$1$, to continue splitting above $n_0$ until some much larger
stage $n_1$. At that level, we again kill off half of them.
Namely, any node beginning with $10$ will become isolated at
$n_1$, and the others will continue splitting up until the very
much later stage $n_2$. And so on.

The general procedure is: at level $n_k$, the nodes beginning with
$1^k0$ will become isolated at level $n_k$ (which is much larger
than $k$), and the nodes beginning with $1^k$ are allowed to
continue branching up to level $n_{k+1}$.

The resulting tree $T$ will have only countably many branches,
since the only branches will be the isolated branches that we
forced to occur, plus the all $11111\cdots$ branch, since whenever
a sequence has its first $0$ in position $k$, it will become
isolated at level $n_k$.

But now, the main point, is that by choosing the levels
$n_0<n_1<n_2<\cdots$ to be very fast growing, we can ensure a
growth rate exceeding $f$. Specifically, since $f$ is 
$o(2^n)$, there is for each $k$ a number $n_k$ such that
$f(n))<2^{n-k}$ for all $n\geq n_k$, and we may also assume
$n_0<n_1<n_2<\cdots$. Now, with the $T$ as defined above, it
follows that the number of nodes of $T$ on level $n_k$ is at least
$2^{n_k-k}$, and it remains at least $2^{n-k}$ for $n_k\leq n<n_{k+1}$, and
this is larger than $f(n)$, as desired. So the growth rate of $T$
is at least $f$, even though $T$ has only countably many paths.
QED