TITLE: Why does a full binary tree of $n$ leaves have $2n-1$ nodes?
QUESTION [22 upvotes]: A complete binary tree is defined as a tree where each node has either $2$ or $0$ children.
A variety of sources have described the relation between nodes and leaves to be $2n-1$ where $n$ is the number of leaves. I haven't however been able to find a description of how this relation was derived.

REPLY [2 votes]: A graph follows a basic property

$$\text{Sum of degrees of all nodes }= 2 \times\text{number of edges}$$

Now since it is complete binary tree it follows:

Leaves are the only nodes with degree $1$; let there be $n$ of them
Root is the only node with degree $2$, there is $1$ root
All internal nodes have degree $3$; let there be $x$ of them

Since tree is also a graph so using the basic property you get
$$n\times 1 + 1\times2 + x\times3 = 2 \times\text{number of edges}$$
To compute the number of edges:
Consider every node except root node, it has exactly one parent edge and every edge is some parent edge. So total number of edges is the number of parent edges which is equal to the number of nodes that have a parent i.e. $n+x$ (root node has no parent)
Hence the equation becomes
$$n + 2 + 3x = 2\times(n+x)$$
Whence
\begin{align}x = n-2\\
\text{Total number of nodes} &= n + 1 + x &+1\text{ is for root node}\\
&= n + 1 + n-2 = 2n-1\end{align}