 Shape of minimum-cost positive-weight ternary prefix trees

For a finite source alphabet with at least two symbols and strictly positive weights, every ternary prefix code minimizing weighted codeword length has all internal vertices of outdegree three, except possibly one of outdegree two whose tree children are deepest tree leaves.

To prove this, let $L$ be the greatest tree leaf depth. A missing tree child at any depth $n<L$ would permit moving a deepest tree leaf into that vacant branch, strictly reducing its positive weighted cost. Thus all internal vertices above depth $L-1$ have three tree children. Tree parents at depth $L-1$ have only tree leaves as tree children; one tree child would permit contraction and strict improvement. If two such tree parents each had two tree children, move one tree child of the second into the first tree parent's third slot, preserving its depth, and contract the second tree parent's remaining tree child. This again strictly improves cost. There is therefore at most one binary tree parent. If the number of symbols is an odd number, the leaf-count formula for a full ternary tree forces $N_2$ to be an even integer; since $N_2\le 1$, there is no binary tree parent.

Strict positivity matters. For probabilities $(1,0,0)$, the prefix code $(0,10,110)$ minimizes expected codeword length at one trit, yet has unary vertices and a deepest tree leaf without a sibling codeword. Removing or rearranging zero-weight branches can select a reduced minimizer, but arbitrary minimizers need not have the positive-weight shape.