the following question came to my mind. Let a tree T be a directed graph with a distinguished root node r, such that there exists a unique path from r to each node v. Moreover, let T(n) denote the subgraph induced by the nodes of depth <=n.
For any two trees T and T', if T(n) and T'(n) are isomorphic for each n, then T and T' are isomorphic as well.
This appears entirely obvious, but I only found a proof for finitely branching trees.
In general, it seems promising to build up a sequence of isomorphisms f_n:T(n)->T'(n) such that f_(n+1) extends f_n for each n. Then the f_n could be extended to an isomorphism of T and T'. But how do I find the f_n?