What is the total number of nodes in a full k-ary tree, in terms of the number of leaves?

Question

I am doing a unique form of Huffman encoding, and am constructing a k-ary (in this particular case, 3-ary) tree that is full (every node will have 0 or k children), and I kn

Answer 1

The formula for 2L-1 that you mentioned comes from looking on a full, complete and balanced binary tree: on the last level you have 2^h leafs, and on the other levels: 1+2+4+....+2^(h-1) = 2^h -1 leafs. When you "mess" levels in the tree and create an unbalanced one, then the number of internal nodes that you have doesn't change.

In 3-ary tree its the same logic: on the last level you have 3^h leafs, and on the other levels: 1+3+9+....+3^(h-1)= (3^h -1 )/2, that means that on a 3-ary tree you have 1.5*L - 0.5 leafs (and this make sence- because the degree is larger you need less internal nodes). I thing that also here, when you mess up levels in the tree you will still need the same number of internal nodes.

Hope that it helps you

Answer 2

For any k-ary tree the total number of nodes n = [(k^(h+1))-1]/(h-1) where h is the height of the k-ary tree.

Ex:- For complete binary tree(k=2) total no. of nodes = [(2^(h+1))-1]/(h-1).

So for height 3 the total no. of nodes will be 15.

For complete ternary tree tree(k=3) total no. of nodes = [(3^(h+1))-1]/(h-1).

So for height 3 the total no. of nodes will be 40.

Answer 3

Think about how to prove the result for a full binary tree, and you'll see how to do it in general. For the full binary tree, say of height h, the number of nodes N is

N = 2^{h+1} - 1

Why? Because the first level has 2^0 nodes, the second level has 2^1 nodes, and, in general, the kth level has 2^{k-1} nodes. Adding these up for a total of h+1 levels (so height h) gives

N = 1 + 2 + 2^2 + 2^3 + ... + 2^h = (2^{h+1} - 1) / (2 - 1) = 2^{h+1} - 1

The total number of leaves L is just the number of nodes at the last level, so L = 2^h. Therefore, by substitution, we get

N = 2*L - 1

For a k-ary tree, nothing changes but the 2. So

N = 1 + k + k^2 + k^3 + ... + k^h = (k^{h+1} - 1) / (k - 1)

L = k^h

and so a bit of algebra can take you the final step to get

N = (k*L - 1) / (k-1)