1. Trees

2. Tree is connected graph without any circuits or cycles
Ex. Trees with one, two, three and four vertices are
A tree must have at least one vertex.
A tree is a simple graph

3. Properties
There is one and only one path between every pair of vertices in a tree
Proof Let T be a tree
 T is a connected graph
There exists at least one path between every pair of vertices in T.
Suppose that there are two distinct paths between two vertices a and b of T.
Union of these two paths will contain a circuit, which is a contradiction that T is a tree.
Hence the theorem.

4. Theorem
If in a graph G, there is one and only one path between every pair of vertices, then G is a tree.
Proof
Since there exists a path between every pair of vertices in graph G.
 G is connected.
Suppose G is not a tree
There is a circuit in G
There are at least two vertices a and b in G, having two distinct paths
But it is given that there is one and only one path between every pair of vertices in G.
Hence the graph is a tree.

5. Theorem
A tree with n vertices has (n - 1) edges
Proof
Let T be a tree with n vertices.
T is a connected graph.
To prove that T has (n - 1) edges.
This theorem will be proved by induction method.
For n = 1,
T is a tree with one vertex.
no. of edges = 0
the theorem is true for n = 1.
For n = 2,
T is a tree with two vertices.
no. of edges = 1
 The theorem is true for n = 2.

6. Suppose the theorem is true for n = k
T, a tree with k vertices has (k – 1) edges.
To prove the theorem is true for n = k + 1
Let e be an edge between two vertices v & w of T
If e is deleted, there is no path between v and w.
[∵ e is a unique path between v & w]
The sub-graph (T – e) will have two components T1 and T2 with k1 and k2 vertices respectively, where
k1 + k2 = k + 1 ; k1, k2  k.

7. ∵ T1 is a tree with k1 vertices.
 T1 has (k1 - 1) edges.
Also T2 is a tree with k2 vertices.
 T2 has (k2 - 1) edges.
Number of edges in T1 and T2
= (k1 - 1) + (k2 - 1) = k1 + k1 – 2 = k + 1 – 2
= k - 1
Thus, the number of edges in T
= (k - 1) + 1 = k
The theorem is true for n = k + 1.
Hence by induction method, the theorem is true for every positive integer n.

8. Theorem
Any connected graph with n vertices and (n - 1) edges is a tree.
Proof
Suppose the connected graph T with n vertices and (n - 1) edges is not a tree.
T has an edge e that is not a bridge.
If e is deleted, the subgraph T – e is still a connected graph with n vertices and (n - 2) edges.

9. We continue this process of locating edges that are not bridges and deleting them until we get a connected subgraph T’ with n vertices and (n - k) edges, where k  1 in which every edge is a bridge.
Now, T’ is a tree with n vertices.
T’ has (n - 1) edges.
Thus,
n – 1 = n – k where k  1
k = but k  1
Which can not be true.
Hence T is a tree.

10. Theorem
In any tree (with two or more vertices), there are at least two pendant vertices.
Proof
Let T be a tree with n vertices.
T has (n - 1) edges.
Sum of degrees of n vertices in T = 2 (n - 1)
Suppose di is the degree of ith vertex
where i = 1, 2, …, n
d1 + d2 + … + dn = 2n – 2

11. Distance If degree of each vertex is more than 1, then
sum of the degrees of n vertices is at least 2n.
But the sum of degrees of all the vertices = 2n – 2.
There are at least two vertices with degree 1.
[∵ no vertex can be of degree zero]
Hence there are at least two pendant vertices.
Distance
In a connected graph G, the distance d(v1, v2) between two vertices v1 and v1 is the length of the shortest path between them.

12. Rooted Trees
A tree in which one vertex (called the root) is distinguished from all the others is called a rooted tree.
Rooted trees with four vertices
Non-rooted trees are called free trees or simply trees.

13. Binary Tree
A tree with n vertices (n  3) is called binary tree in which there is exactly one vertex of degree two, each of the remaining vertices is of degree one or three.
The vertex of degree 2 is called the root as it is distinct from all other vertices, it serves as a root.
Binary tree is always a rooted tree.
Non-pendant vertex is called Internal Vertex

14. Properties The number of vertices in a binary tree is always odd Proof
Let n be the number of vertices in a binary tree.
Since there is exactly one vertex of even degree.
 All the remaining (n -1) vertices are of odd degrees.
Since the number of odd degree vertices in a graph is always even.
(n - 1) is even.
Hence n is odd.

15. Theorem
Number of pendant vertices in a binary tree is
(n+1)/2
Proof
Let T be the binary tree of n vertices and p be the number of pendant vertices in T.
Also there is one vertex of degree two in T.
n – p – 1 is the number of vertices of degree three
Sum of degrees of all vertices in T
= p + 3(n – p - 1) + 2
= 3n – 2p – 1
Since sum of degrees of all vertices in a graph is twice the number of edges.

16. Number of edges in T = ½(3n – 2p - 1)
Also, number of edges in a tree of n vertices = n – 1
½(3n – 2p - 1) = n – 1
p = (n + 1)/2
The number of internal vertices in a binary tree is one less than the number of pendant vertices.
In a binary tree, a vertex vi is said to be at level li if vi is at a distance of li from the root.

17. Height of Tree
The maximum level, lmax in a binary tree is called the height of the tree.
Minimum possible height of an n-vertex binary tree is
min lmax =  log2(n + 1) - 1
Maximum possible height of an n-vertex binary tree is max lmax = (n - 1)/2.

18. Path Length Also called external path length of a tree
= Sum of the path lengths from the root to all
pendant vertices.
The path length of a binary tree is often directly related to the execution time of an algorithm.
A binary tree with minimum possible height gives the minimum path length for a given n.