y \rightarrow s^{-}$ does not hold then it is immediately clear that $y$ cannot be put on side $s$. So assume (conversely) that this condition does hold, and for the same reason assume there is a split cluster $C$ for side $s$ that contains $y$. In other words, there is a cluster $C$ such that $\{ y, s^{+}, s^{-} \} \cap C = \{ y, 
s^{-} \}$. Since $t^{+} \rightarrow y \rightarrow t^{-}$ holds, it follows that $C$ also contains $t^{+}$ and $t^{-}$, because any cluster that contains $y$ also contains $t^{-}$, and we know that $C$ contains either both of $t^{+}$ and $t^{-}$, or neither of them. However, there is no edge in $N$ that can represent $C$: the only edges that represent $C$ lie on side $s$, but the fact that $s$ is the lowest side means that no cluster beginning on side $s$ can contain any leaves on side $t$. To summarize, we have a simple test for determining whether a leaf should be placed on side $s$. Once we have determined the set of leaves that should be placed on side $s$, it is easy to determine the correct order of those leaves by inspecting $\rightarrow 