data structurediscrete mathematicsgraph theoryinductiontrees
Given a tree $T$ with $n$ nodes, every node that isn't a leaf has 2 children, can i drive a formula for number of leaves in a such tree $T$?
My idea: i read this post:Full binary tree proof validity: Number of leaves $L$ and number of nodes $N$
, but it is about full binary tree, our problem maybe not full binary tree.
Given h...height if tree, N(h).. count of nodes for tree height h. If h = 1: N(h) = 1; h = 2: N(h) = N(1) * 2 = 1 * 2; h = 3: N(h) = N(2) * 2 = N(1) * 2 * 2 = 1 * 2 * 2 * 2; ... h = n: N(n) = N(n-1) * 2 = ... = 1 * 2 * 2 * 2...= 1 * 2^n = 2^n Each node has two children.
So, count of nodes if height of tree is n is N(n) = 2^n. And if count of nodes of full binary tree is N, then height of tree n is proportional to log_2(N) or n = C(log(N)).
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
Suppose that $T$ has $n$ nodes, $\ell$ of which are leaves; then the sum of the degrees of the nodes is
$$2+3(n-\ell-1)+\ell=3n-2\ell-1\,,$$
since the root has degree $2$, each of the other $n-\ell-1$ non-leaves has degree $3$, and each leaf has degree $1$. This is twice the number of edges, so $T$ has $\frac{3n-1}2-\ell$ edges.
On the other hand, $T$ is a tree, so it has $n-1$ edges, and we have the equation
$$\frac{3n-1}2-\ell=n-1\,,$$
so that
$$\ell=\frac{n+1}2\,.$$
The reason that $n$ has to be odd, as this formula clearly implies, is that the non-root nodes of $T$ come in sibling pairs, so there is an even number of them, and the root makes an odd total number of nodes.