Perfect ternary tree ($N_h=(3^{h+1}-1)/2$)

A full ternary tree whose leaves all have the same depth. At depth $j$ there are $3^j$ vertices, so a tree with $h+1$ levels has $N_h={\sum }_{j=0}^h{3}^j=(3^{h+1}-1)/2$ vertices and $3^h$ leaves, by the finite geometric series formula. For a rose tree representation, recursively flattening children and joining their lists with ordinary list concatenation copies each label once per ancestor. The work is therefore $\Theta ((h+1)3^h)=\Theta (N_h\log N_h)$ when the complete output is demanded. Passing an already constructed suffix through the recursion instead emits each label once, giving $\Theta (N_h)$ work.