Ternary heap is like a binary tree, just every node can have up to $3$ sons and not $2$.

I try to bound the number of nodes in the heap, $n$, using the height of the heap $h$.

The solutions get to:

$$ 3^h < n < 3^{h+1} $$

Yet, I get to:

$$ \frac{3^h}{2} < n < \frac{3^{h+1}}{2} $$

In short, what I do is:

$$ \sum_{i = 0}^{h - 1}3^i = \frac{3^h-1}{2} $$

For all the nodes in all levels except for the last level, because all the levels are full.

If the last level is full:

$$ \frac{3^h - 1}{2} + 3^h $$

If the last level has only $1$ node, we get:

$$ \frac{3^h - 1}{2} + 1 $$

From here I conclude what I showed at the beginning.

Why the solutions get to something else?


The expected solution is wrong.

Pick $h=0$. The only ternary heap with height $0$ has only one node. Nevertheless the expected solution says that it needs to have at least $3^0 + 1 = 2$ nodes.

In case you meant to write $3^h \le n \le 3^{h+1}$ instead of $3^h < n < 3^{h+1}$, the expected solution is still wrong. Consider a ternary heap with $n=2$ nodes. This heap has height $1$ but according the expected solution it should have at least $3^h = 3$ nodes.

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  • $\begingroup$ So the height of a ternary tree is $ h = \left \lfloor log_32n \right \rfloor$? $\endgroup$ – Alon May 27 at 19:52
  • $\begingroup$ Yes.$\phantom{}$ $\endgroup$ – Steven May 27 at 20:13

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