Ordinary induction - Number of nodes in a full binary tree

Question

Question: Use ordinary induction to show the minimum number of nodes for a full binary tree of height h.

I started to play with it and I believe I got the answer $2^{h+1}-1$. However, I am not able to write the proof by induction.

And another small question, I believe that in this case the minimal and maximal amount of nodes are similar (since it's a full binary tree). Am I wrong?

Thanls

You could review this answer for how to think of an induction proof. Yes, the number of nodes is fixed. — Ross Millikan, Dec 07 '19 at 17:29
You add some leaves when you obtain tree of height $h+1$ form tree of height $h$. How many? That's the induction step. — Alexey Burdin, Dec 07 '19 at 17:29

score 1 · Answer 1 · answered Dec 07 '19 at 17:55

Yes, the use of 'minimum' here seems odd since a full tree has precisely $2^{h+1}-1$ nodes as you say.

The inductive step would be as follows.

Assume a tree of height $k$ has $2^{k+1}-1$ nodes. Then a tree of height $k+1$ has $(2^{k+1}-1) + 2^{k+1}$ nodes. This equals $2^{k+2}-1$ nodes as required.

Ordinary induction - Number of nodes in a full binary tree

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