 Solution

Assume a full-support finite distribution. The preceding structural arguments show that every internal vertex has three tree children, except possibly one tree parent at depth $L-1$ with two tree children. Let $N_3$ and $N_2$ count these two kinds of internal vertex. Part iv gives $N_2\in \{0,1\}$; no vertex has one tree child.

A finite rooted tree with $m$ tree leaves has $m+N_2+N_3-1$ edges, while counting outgoing edges gives $2N_2+3N_3$. Equating the counts yieldsFor an odd integer $m$, the right side forces $N_2$ to be an even integer. Since it is at most one, $N_2=0$. Thus no deepest pair has only two sibling codewords, and the optimal code tree is a full ternary tree. This leaf-count identity also directly proves that a full ternary tree has an odd number of tree leaves.

Again the unrestricted zero-weight version is false: $(0,10,11)$ is an optimal prefix code for $p=(1,0,0)$, with odd integer $m=3$, but its two deepest tree leaves form a two-sibling pair. There is no contradiction with the proof because its tree root has only two tree children, which the full-support finite distribution assumption would forbid above the deepest level.