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Here is the description of a proof problem:

A complete binary tree of depth N is a binary tree in which every node on levels $0,1,2,...,N-1$ is a parent and has two children, and each node on level N is a leaf.

It's asking for proving binary tree of depth N has $2^{N+1} - 1$ nodes.

I am not really sure how to approach this proof. I tried to plug in some values in $D(N) = 2^{N+1} - 1$ just to play with the formula, I get $D(1) = 2^{1+1} - 1 = 3$, $D(2) = 2^{2+1} - 1 = 7$, $D(3) = 2^{3+1} - 1 = 15$, and so on. I still did not see how these numbers relate to the descriptions given and how I would relate these values to given facts to build a proof.

Romantic Electron
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Nate Lee
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  • Your claim is true if only the tree is a perfect binary tree. A complete tree may have fewer than $2^{N+1}-1$ nodes. A perfect binary tree is a binary tree in which all interior nodes have two children and all leaves have the same depth or same level. – fade2black Sep 03 '17 at 18:09
  • Related question. Also this. Duplicate? – Raphael Sep 03 '17 at 18:11
  • @fade2black There are multiple competing definitions of "complete binary tree". Unfortunately, the definition in the question doesn't work: a path is a "complete binary tree" by this definition, since it doesn't require interior nodes to have two children. – David Richerby Sep 03 '17 at 19:01
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    @fade2black Correct: the given definition doesn't imply any specific number of nodes. As I said, a path fits the definition and has only $N$ nodes. – David Richerby Sep 03 '17 at 19:13
  • Your definition is missing the following part: each node on levels $1,2,\ldots,N-1$ is a parent and has two children, and each node on level $N$ is a leaf. – Yuval Filmus Sep 03 '17 at 21:04
  • @YuvalFilmus, I overlooked the two children part there. But doesn't having two children makes the complete binary tree a full binary tree? This has gotten me quite confused since I use this source. – Nate Lee Sep 03 '17 at 22:09
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    Well, it's clearly the right definition here. As for the concepts complete and full, they are not as standard as some people here seem to think, and might mean slightly different things to different people. – Yuval Filmus Sep 03 '17 at 22:16

1 Answers1

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Hint 1: Draw some binary trees of depth $0, 1, 2$ and $3$. Depth $0$ is only the the root.

Hint 2: Use Induction on the depth of the tree to derive a proof.

The base case is depth $n=0$. With depth $0$ we only have the root, that is, $2^{0 + 1} - 1 = 1$ nodes, so the formula is valid for $n=0$.

The next case is depth $n=0+1=1$, since the tree is full the root must have $2$ leafs. So we have $1 + 2 = 3$ nodes in total. The formula still holds since $2^{1+1} - 1 = 3$.

Can you generalize this procedure? Does this hold for all $n \geq 0$?

Aristu
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