trees
If the level ($\lambda$) of a full binary tree at zero is just a root node, than I know that I can get the maximum possible number of nodes (N) for a full binary tree using the following:
N = $2^{\lambda+1}$- 1
Is the minimum possible number of nodes the following?
N = 2*$\lambda$ + 1
Hint You want a minimum node depth $d$ balance binary search tree. What happens if you take a maximum node tree of depth $d-1$ and add a single node?
My preferred definition of a non-empty, full binary tree is:
Definition 0. A pointed magma is an ordered triple $(X,\bullet,\star)$, such that:
- $X$ is a set
- $\bullet$ is an element of $X,$ and
- $\star$ is a function $X \times X \rightarrow X$.
Definition 1. Write $\mathbb{T}$ for the initial pointed magma, the elements of which are called non-empty full binary trees.
For example, here's an example of a non-empty full binary tree:
$$(\bullet \star \bullet) \star \bullet \in \mathbb{T}$$
This can be visualized as a tree, of course:
Okay. Let $\varphi : \mathbb{T} \rightarrow \mathbb{N}$ denote the node-counting function. Explicitly:
Definition 2. Define a function $\oplus : \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$ as follows: $$a\oplus b = a+b+1$$
Definition 3. Let $\varphi : (\mathbb{T},\bullet,\star) \rightarrow (\mathbb{N},1,\oplus)$ denote the unique such homomorphism of pointed magmas. We call $\varphi$ the node counting function.
For example, $$\varphi((\bullet \star \bullet) \star \bullet) = (1\oplus 1) \oplus 1 = 3 \oplus 1 = 5$$
So the above tree has $5$ nodes, as desired.
I claim that:
Claim. For all $x \in \mathbb{T}$, the natural number $\varphi(x)$ is odd.
This prove this, we need a way of performing induction on non-empty full binary trees. Here's a theorem that lets us do this:
Structural Induction for $\mathbb{T}$.
The pointed magma $(\mathbb{T},\bullet,\star)$ has no proper subalgebras.
More explicitly:
Structural Induction for $\mathbb{T}$. (Long form.)
Let $X$ denote a subset of $\mathbb{T}$.
If
- $\bullet \in X$, and
- for all $x,y \in X$, we have $x \star y \in X$,
then
$X = \mathbb{T}$.
Now we can prove our claim. Let $X$ denote the set of all $x \in \mathbb{T}$ such that $\varphi(x)$ is odd.
We know that $\bullet \in X$, since $\varphi(\bullet) = 1$ and $1$ is odd.
Assume $x,y \in X$. We will show that $x \star y \in X$. We know that $\varphi(x \star y) = \varphi(x) \oplus \varphi(y) = \varphi(x)+\varphi(y)+1.$ But the sum of three odd numbers is itself odd. So $\varphi(x \star y)$ is odd.
Hence by the structural induction theorem, we have $X=\mathbb{T}$. But recall that we defined $X$ as the set of all $x \in \mathbb{T}$ such that $\varphi(x)$ is odd. Thus $\mathbb{T}$ itself is the set of all $x \in \mathbb{T}$ such that $\varphi(x)$ is odd. In other words, $\varphi(x)$ is always odd, for all $x \in \mathbb{T}$. QED
A further comment.
It is worth noting that we can define the leaf-counting function in much the same way:
Definition 4. Let $\lambda : (\mathbb{T},\bullet,\star) \rightarrow (\mathbb{N},1,+)$ denote the unique such homomorphism of pointed magmas.
We call $\lambda$ the node counting function.
For instance, $$\lambda((\bullet \star \bullet) \star \bullet) = (1+1)+1 = 3,$$ so the aforementioned tree has $3$ leaves, as desired.
The cool thing about $\lambda$ is that it gives what is perhaps the most fundamental definition of the Catalan numbers; namely, writing $C_n$ for the $n$th Catalan number, we can define: $$C_n = |\lambda^{-1}(n+1)|$$
For example, $$C_2 = |\lambda^{-1}(3)| = |\{(\bullet \star \bullet) \star \bullet, \bullet \star (\bullet \star \bullet)\}| = 2$$
Best Answer
For a full binary tree $T$ of height $\lambda$, I believe that the maximum number of nodes is $N = 2^{\lambda + 1} - 1$ (not $+1$.)
It seems likely that you can prove the minimum number of nodes for a full binary tree of height $\lambda$ inductively. (We can readily verify that the minimum number of nodes for $\lambda = 1$ is $2\times 1 + 1 = 3$, showing the base case to be true.) Assuming (inductively) that for $\lambda = k$ we have a minimum of $N = 2k+1$ nodes, if we add a node, it must branch from one of the leaves. But in order to maintain a full binary tree, we must add an additional node; that is, adding an additional levels requires at minimum two more nodes. So, we will have $N+2$ nodes. Then, by our induction hypothesis $N+2 = (2k+1) + 2 = 2(k+1) + 1$, which is what we wanted.
Not exactly formal, but does that make sense?