I need to prove that for every full binary tree $T$(a tree in which every node other than the leaves has two children) with $n$ leaves there exists a sequence of probabilities $f_1, f_2...f_n$ that one of its huffman tree is $T$.
I thought to prove this by induction on the depth of $T$, and take the sequence of $f_n=\frac{1}{2^d}$. For depth $d=0$ we have $f_n=1$ and the only huffman tree that suits for this sequence is $T$.
I don't know how to procceed from here.