Suppose you have two rooted trees $T\_1$ and $T\_2$ with roots $r\_1$ and $r\_2$, respectively. Furthermore, for every $k\ge 0$, the number of walks of $T\_1$ starting at $r\_1$ of length $k$ is equal to the number of walks of $T\_2$ starting at $r\_2$ of length $k$. Is it true that $T\_1$ and $T\_2$ are isomorphic as rooted trees? 


It seems to me strange that this would be true and I cannot find any mention of it. On the other hand, I can't seem to come up with a counterexample.


EDIT: Apologies! I should have said that walks are not assumed to be simple or closed! A walk is just a sequence of adjacent edges, edges and vertices may be repeated and the ending point does not matter.



A criterion for rooted trees to be isomorphic based on walks
Take a 9-cycle with vertices $\{0,\ldots,8\}$ and join new vertices of degree one to 0, 3, 6. Delete the vertex 8 from the cycle, producing a tree $T$ on 11 vertices. Then $T\backslash 2\cong T\backslash 5$, but no automorphism of $T$ maps 2 to 5. 


Now there is a theorem that if $u$ and $v$ are vertices in graph $X$ and $X\backslash u$ and $X\backslash v$ are cospectral and their complements are cospectral, the generating functions for closed walks at $u$ and for all walks at $u$ are equal to the corresponding generating functions for
walks at $v$. (This is a consequence of results from chapter 4 in my book "Algebraic Combinatorics".) 


So vertices 2 and 5 cannot be distinguished by your criterion. (My example is quite likely not minimal.)


