 Solution

Let $L$ again denote the largest codeword length, and assume a full-support finite distribution. The missing-branch exchange from part iii forces every internal vertex at depth at most $L-2$ to have three tree children. At depth $L-1$, all tree children are tree leaves. Part ii rules out one tree child there; the hypothesis that there is no pair of two deepest sibling codewords rules out two tree children. Thus each such tree parent also has three tree children. Every internal vertex has exactly three tree children, which is a full ternary tree, called semi-complete in this paper. The same last-level argument covers $L=1$.

Without the assumption of a full-support finite distribution the implication is false: for $p=(1,0,0)$, the optimal prefix code $(0,10,110)$ has no deepest pair of two sibling codewords, but its code tree contains vertices with a single tree child. Empty unused tree children are not vertices of the code tree.