A vertex $v$ of a connected graph $G$ is said to be a boundary vertex of $G$ if for some other vertex $u$ of $G$, no neighbor of $v$ is further away from $u$ than $v$. The boundary $\partial(G)$ of $G$ is the set of all of its boundary vertices. The boundary distance matrix $\hat{D}_G$ of a graph $G=([n],E)$ is the square matrix of order $κ$, being $κ$ the order of $\partial(G)$, such that for every $i,j\in \partial(G)$, $[\hat{D}_G]_{ij}=d_G(i,j)$. Given a square matrix $\hat{B}$ of order $κ$, we prove under which conditions $\hat{B}$ is the distance matrix $\hat{D}_T$ of the set of leaves of a tree $T$, which is precisely its boundary. We show that if $G$ is either a block graph or a unicyclic graph, then $G$ is uniquely determined by the boundary distance matrix $\hat{D}_{G}$ of $G$ and we also conjecture that this statement holds for every connected graph $G$, whenever both the order $n$ and the boundary (and thus also the boundary distance matrix) of $G$ are prefixed. Moreover, an algorithm for reconstructing a 1-block graph (resp., a unicyclic graph) from its boundary distance matrix is given, whose time complexity in the worst case is $O(κn)$ (resp., $O(n^2)$).