After naming the nodes of a Binomial expanding tree as follows: 
I want to find a way to sequentially iterate through the nodes of a tree with this structure, but one with indefinite branch lengths.
What formula can I use (Algebraically or in Code) to determine the name of the next node in this sequence, based only on the name of the previous Node.
Here is the start of the sequence of Node Names:
[0], [1], [2], [1,1], [3], [2,1], [1,2], [1,1,1], [4], [3,1], [2,2], [1,3], [2,1,1], [1,2,1], [1,1,2], [1,1,1,1], [5], [4,1], [3,2], [2,3], [1,4], [3,1,1] ...
What I have so far:
If all values in the name of the previous node equal 1, then the next node name equal one value equal to the sum of the values of the previous node, plus one.
Examples: [1] => [2], [1,1,1] => [4]
If all values, except the last one, in the name of the previous node equal 1, then the next node name starts with the sum of the values of the previous node minus the amount of values in the previous node name. This starting number is then followed by ones until the sum of this node name equals the sum of the previous node.
Examples: [3] => [2,1], [1,1,3] => [2,1,1,1], [6] => [5,1], [1,4] => [3,1,1]
Now I am stuck the remaining node names.
Note: The amount of values (Length) and the sum of the values (Total) of these node names do not change as you move from the previous node to the next one, you only iterate through the possible permutations.
Not exactly the sequence you have given, but close:
$$ \operatorname{next}\big((a_0,a_1,\ldots,a_{n-1})\big) = \Big(a_0,a_1,\ldots,a_{k-2}, a_{k-1} - 1, 1+\sum_{i = k}^{n-1} a_i\Big) $$ where $$k = \min\Big\{j \in \mathbb{N} \ \Big|\ \forall i \in \{j,\ldots,n-1\}.\ a_i = 1 \Big\} $$ under convention in which sequences \begin{align} &(a_0, a_1, \ldots, a_{k-2}) \\ &(a_0, a_1, \ldots, a_{k-1}, a_{k-1}-1) \\ \end{align} and set $\{j, \ldots, n-1\}$ are empty for $k=1$, $k=0$ and $j > n-1$ respectively.
Starting from $(1)$ we get following tuples: \begin{align} &(1) \\ &(2) \\ & &&(1, 1) \\ &(3) \\ & &&(2, 1) \\ & &&(1, 2) &&(1, 1, 1) \\ &(4) \\ & &&(3, 1) \\ & &&(2, 2) &&(2, 1, 1) \\ & &&(1, 3) &&(1, 2, 1) &&(1, 1, 2) &&(1, 1, 1, 1) \\ &(5) \\ & &&(4, 1) \\ & &&(3, 2) &&(3, 1, 1) \\ & &&(2, 3) &&(2, 2, 1) &&(2, 1, 2) &&(2, 1, 1, 1) \\ & &&(1, 4) &&(1, 3, 1) &&(1, 2, 2) &&(1, 2, 1, 1) &&(1, 1, 3) \\ & && &&(1, 1, 2, 1) &&(1, 1, 1, 2) &&(1, 1, 1, 1, 1) \end{align}
I hope this helps $\ddot\smile$