 Solution

At the root all $z$ neighbours are in shell one, so the probability of leaving the root for that shell is one. Conversely, a walker anywhere in shell one chooses its unique rootward edge with probability $1/z$. The two boundary recurrences are thereforeTogether with $P(0,0)=1$ and $P(r>0,0)=0$, these equations determine the distribution. In particular $P(1,1)=1$ and $P(0,2)=1/z$, which check the special root coefficient. For the original indexing, replace $N+1$ by $N$ and $N$ by $N-1$ on the right-hand sides. Summing these equations and the bulk recurrence conserves ${\sum }_{r\ge 0}P(r,N)=1$.