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Is the root node an internal node?

So I've looked around the web and a couple of questions here in stackoverflow here are the definition:

  • Generally, an internal node is any node that is not a leaf (a node with no children)
  • Non-leaf/Non-terminal/Internal node – has at least one child or descendant node with degree not equal to 0
  • As far as i understand it, it is a node which is not a leaf.

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?

  • As the wonderful picture shows, internal nodes are nodes located between the root of the tree and the leaves

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?

like image 786
Programmerboi Avatar asked Jan 18 '13 04:01

Programmerboi


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1 Answers

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
like image 190
Utkarsh Avatar answered Oct 15 '22 11:10

Utkarsh