What is the oxidation state of nitrogen in hydroxylamine?

The oxidation number of $N$ in hydroxylamine, ${H}_{2} N - O H$, is $- I$.
By definition, the oxidation number of any atom in a molecule, is the charge left on that atom, when all the bonding pairs of electrons are conceptually broken, with the charge going to the most electronegative atom. Break a $N - H$ bond and we get ${N}^{-}$ and ${H}^{+}$. Break a $N - O H$ bond and we get ${N}^{+}$, and $H {O}^{-}$, because oxygen is more electronegative than nitrogen, the oxygen centre gets the electron (please note that this is entirely a conceptual exercise!).
The sum of the individual oxidation numbers must equal the charge on the molecule or the ion. Here, we have a neutral molecule. So $3 \times O {N}_{H} + O {N}_{O} + O {N}_{N}$ $=$ $0$; so $3 \times 1 - 2 + O {N}_{N} = 0$. Thus $O {N}_{N} = - I$.
Can you hazard a guess (or a reasoned answer) as to the formal oxidation numbers of nitrogen in ammonia, $N {H}_{3}$, and in hydrazine, ${H}_{2} N - N {H}_{2}$?