inductiontrees
How would I prove that any binary tree that has n leaves has precisely $2n-1$ nodes ?
Given that a binary tree is either a single node "o" or a node with the left and right subtrees contains a binary tree
o
/ \
binary binary
tree tree
Any help would be appreciated !
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, $$C_2 = |\lambda^{-1}(3)| = |\{(\bullet \star \bullet) \star \bullet, \bullet \star (\bullet \star \bullet)\}| = 2$$
Hint:
I hope this helps $\ddot\smile$
Best Answer
Your formula only works if you assume all the leaves are the same depth in the tree and every node that isn't a leaf has 2 children (see wikipedia for different kinds of binary trees). For example imagine a tree
This has $n=1$ leaves and 2 nodes but the formula gives $2n-1=1$.
Making this assumption, to prove by induction, notice (1) that the formula holds true for a tree of height 1 with 1 node, because $2\times 1 - 1 = 1$.
Then (2) assume that the formula holds for trees with $k$ leaves, so assume trees with $k$ leaves have $2k-1$ nodes. Adding another level to the tree with $k$ leaves adds another $2k$ leaves because each leaf in the original tree has 2 children. So this new tree has a total of $2k-1$ leaves from the original plus another $2k$ leaves = $4k-1$ leaves. The formula for $2k$ leaves gives $2(2k)-1=4k-1$ leaves, which is the same!
So because our (1) our base step is true; and (2) our inductive step is true, then the formula is true for all $n$ (subject to the constraint above).
Alternatively, the depth $d$ of the tree is $\log_2 n +1$ because the number of nodes doubles at each level.
The total number of nodes is $1 + 2 + 4 + \ldots$ with $d$ terms in the sum, or $2^0 + 2^1 + 2^2 + \ldots + 2^{d-1}$. By the geometric series formula the sum is equal to
$$ \begin{align}\frac{2^d -1}{2-1} &= 2^{\log_2 (n+1)} - 1 \\&= 2\times2^{\log_2 n} -1\\&= 2n -1\end{align}$$