It is obvious that the total number of the leaves of $S$ and $T$ are equal. We solve the problem using induction. The base case for $2$ leaves is trivial because there is no vertex of degree $2$. Now suppose that $S, T$ have $k$ leaves. Let

$$S' = S - \{u\}, \quad T' = T - \{f(u)\}$$

in which $u$ is a leaf. It means that we omitted leaf $u$ from $S$, and leaf $f(u)$ from $T$. Assume the induction hypothesis holds for $S', T'$, so there is a correspondence between their vertices, like $g$. Now let $u'$ be the neighbor of $u$ in $S'$, and $u''$ be the neighbor of $f(u)$ in $T'$. It is enough that we prove that $u'' = g(u')$ and the length of the edge $uu'$ is equal to the length of the edge $f(u)u''$.

First note that if $v, w$ be two leaves in $S$ and the path between $v, u$, intersects the path between $v, w$ for the first time in vertex $k$, then we have: (let $l(uv)$ denote the length of the path between $u$ and $v$)
$$l(uk) = \frac{l(uv) + l(uw) - l(vw)}{2}$$
And since $S$ doesn't have a $2$-degree vertex:
$$\begin{align*} \text{length of edge } uu' &= \min_{v,w \in S_{\text{leaves}}} \left\{ \frac{l(uv) + l(uw) - l(vw)}{2} \right\} \\ &= \min_{f(v), f(w) \in T_{\text{leaves}}} \left\{ \frac{l(f(u)f(v)) + l(f(u)f(w)) - l(f(v)f(w))}{2} \right\} \\ &= \text{length of the edge } f(u)u'' \end{align*}$$
Now do the same for $T$. Thus, because these two values for a pair $v, w$ and its correspondence $f(v), f(w)$, minimizes, so $u'' = g(u')$ and the length of the edge $uu'$ is equal to the length of the edge $f(u)u''$, hence the induction and proof are complete.