So I've looked around the web and a couple of questions here in stackoverflow here are the definition:
I was about to conclude that the root is also an internal node but there seems to be some ambiguity on its definition as seen here:
What is an "internal node" in a binary search tree?
If we follow that definition then the root node isn't going to be counted as an internal node. So is a root node an internal node or not?
if a tree has only one node then we say that it is a tree with only root node, we never say that the tree has a single leaf node. Since internal node means a non-leaf node and because root node is never considered as leaf node I would say that in case of single node tree root node is an internal node.
The root is an internal vertex unless it is the only vertex in the graph, in which case it is a leaf. Supportive Theorem: For any positive integer n, if T is a full binary tree with n internal vertices, then T has n + 1 leaves and a total of 2n + 1 vertices.
An internal node is any node that has at least one non-empty child. Figure 7.2. 1: A binary tree. Node A is the root.
Definition: A node of a tree that has one or more child nodes, equivalently, one that is not a leaf. Also known as nonterminal node. See also parent, root.
Statement from a book : Discrete Mathematics and Its Applications - 7th edition By Rosen says,
Vertices that have children are called internal vertices. The root is an internal vertex unless it is the only vertex in the graph, in which case it is a leaf.
Supportive Theorem:
For any positive integer n, if T is a full binary tree with n internal vertices, then T has n + 1 leaves and a total of 2n + 1 vertices.
case 1:
O <- 1 internal node as well as root
/ \
O O <- 2 Leaf Nodes
case 2: Trivial Tree
O <- 0 internal vertices (no internal vertices) , this is leaf
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