Pre-requisites: Handshaking theorem.
Pendant Vertices
Let G be a graph, A vertex v of G is called a pendant vertex if and only if v has degree 1. In other words, pendant vertices are the vertices that have degree 1, also called pendant vertex.
Note: Degree = number of edges connected to a vertex
In the case of trees, a pendant vertex is known as a terminal node or leaf node, or leaf since it has only 1 degree. Remember leaf nodes have only 1 degree so pendant vertex is called a leaf node in the case of trees.
Example: In the given diagram A and B are pendant vertices because each of them has degree 1.
Graph 1.0
Let’s take this example to print all the pendant vertices in the graph.
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to print all the pendant// verticesvoid printPendantVertices(map<char, vector<char> > graph){ // All the vectors which contain only 1 // vertex i.e, size 1 has only 1 edge // hence a pendant vertex. for (auto x : graph) { if (x.second.size() == 1) { cout << x.first << " "; } }}// Driver Codeint main(){ map<char, vector<char> > graph; graph['A'].push_back('B'); graph['B'].push_back('A'); graph['C'].push_back('B'); graph['B'].push_back('C'); printPendantVertices(graph); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG { // Function to print all the pendant // vertices static void printPendantVertices( HashMap<Character, ArrayList<Character> > graph) { // All the vectors which contain only 1 // vertex i.e, size 1 has only 1 edge // hence a pendant vertex. for (Character x : graph.keySet()) { if (graph.get(x).size() == 1) { System.out.print(x + " "); } } } public static void main(String[] args) { HashMap<Character, ArrayList<Character> > graph = new HashMap<>(); graph.put('A', new ArrayList<>()); graph.get('A').add('B'); graph.put('B', new ArrayList<>()); graph.get('B').add('A'); graph.put('C', new ArrayList<>()); graph.get('C').add('B'); graph.get('B').add('C'); printPendantVertices(graph); }}// This code is contributed by Karandeep1234 |
Python3
# Python program for the above approach# Function to print all the pendant# verticesdef printPendantVertices(graph): # All the vectors which contain only 1# vertex i.e, size 1 has only 1 edge# hence a pendant vertex. for x,y in graph.items(): if (len(y) == 1): print(x, end = " ")# Driver Codegraph ={}graph["A"] = ["B"]graph["B"] = ["A"]graph["C"] = ["B"]graph["B"].append("A")printPendantVertices(graph)# This code is contributed by shinjanpatra |
C#
using System;using System.Collections.Generic;class GFG{ // Function to print all the pendant vertices static void printPendantVertices( Dictionary<char, List<char> > graph) { // All the vectors which contain only 1 // vertex i.e, size 1 has only 1 edge // hence a pendant vertex. foreach(var x in graph.Keys) { if (graph[x].Count == 1) { Console.Write(x + " "); } } } public static void Main(string[] args) { var graph = new Dictionary<char, List<char> >(); graph['A'] = new List<char>(); graph['A'].Add('B'); graph['B'] = new List<char>(); graph['B'].Add('A'); graph['C'] = new List<char>(); graph['C'].Add('B'); graph['B'].Add('C'); printPendantVertices(graph); }}// This code is contributed by ishankhandelwals. |
Javascript
<script>// Javascript program for the above approach// Function to print all the pendant// verticesfunction printPendantVertices(graph) { // All the vectors which contain only 1 // vertex i.e, size 1 has only 1 edge // hence a pendant vertex. for (x of graph) { if (x[1].length == 1) { document.write(x[0] + " "); } }}// Driver Codelet graph = new Map();graph.set("A", ["B"]);graph.set("B", ["A"]);graph.set("C", ["B"]);graph.set("B", [...graph.get("B")].push("A"));printPendantVertices(graph);// This code is contributed by saurabh_jaiswal.</script> |
A C
Complexity :
Time complexity : O(n), where n is the number of vertices in the graph. This is because the code iterates through all the vertices in the graph once and checks if the size of the vector associated with each vertex is equal to 1.
Space complexity : O(n) as the code uses a map to store the graph, where each vertex is a key and the value is a vector containing its adjacent vertices. The size of the map is directly proportional to the number of vertices in the graph.
Non-Pendant Vertices
Non-Pendant Vertices are the vertices that have degrees other than 1. In the case of trees, a non-pendant vertex is a non-leaf node, since it does not have degree 1 ( leaf nodes have degree 1).
Example: In the given diagram A and C are pendant vertices because A and C have degree 1, B is a non-pendant vertex because it contains degrees other than 1 which is 2.
Graph 2.0
Pendant Edge
An edge of a graph is said to be a pendant edge if and only if one of its vertices is a pendant vertex.
Example: In the given diagram AB is a pendant edge since it contains pendant vertex as one of its vertexes.
Graph 3.0
Non-Pendant Edge
An edge of a graph is said to be a non-pendant edge if it does not contain a pendant vertex as one of its vertexes.
Example: in the given diagram AB is a pendant edge since it has pendant vertex (A) as one of its vertexes. BD, BC, DC are non-pendant vertices.
Graph 4.0
Examples:
Let us see some questions based examples on the above topics:
Q1. Suppose that a tree T has 2 vertices of degree 2, 4 vertices of degree 3, and 3 vertices of degree 4. find the number of pendant vertices in T.
Finding number pendant vertices is nothing but finding the number of leaf nodes.
Let’s use the Handshaking Theorem formulaSum of all degrees = 2 * (Sum of Edges)
(2 vertices) * (2 degrees) + (4 vertices) * (3 degrees) + (3 vertices) * (4 degrees) + (k vertices) * (1 degree) = (2 * edges)
where
k is pendant vertices or leaf nodes which have degree 1
e is total number of edges in the tree2*2 + 4*3 + 3*4 + k*1 = 2*e ——-(1)
remember : number of edges = number of vertices – 1
e=(2+4+3+k)-1
e=9+k-1
e=8+k ——-(2)
putting equation 2 in 1 gives
4 + 12 + 12 + k = 2(8+k)
28 + k = 16 + 2k
-2k + k = 16 – 28
-k = -12
k = 12
so total number of pendant vertices are 12
Q2. If a tree T has 4 vertices of degree 2, 1 vertex of degree 3 and 2 vertices of degree 4 and 1 vertex of degree 5. find the number of pendant vertices in T.
Finding number pendant vertices is nothing but finding the number of leaf nodes.
Let’s use the Handshaking Theorem formula
Sum of all degrees = 2 * (Sum of Edges)
(4 vertices)*(2 degrees) + (1 vertex)*(3 degrees) + (2 vertices)*(4 degrees) + (1 vertex)*(5 degree) + (k vertices)*(1 degree) = (2 * edges)
where
k is pendant vertices or leaf nodes which have degree 1
e is total number of edges in the tree4*2 + 1*3 + 2*4 + 1*5 + k*1 = 2*e ——-(1)
remember number of edges = number of vertices – 1
e=(4+1+2+1+k)-1
e=8+k-1
e=7+k ——-(2)
putting equation 2 in 1 gives
8 + 3 + 8 + 5 + k = 2(7+k)
24 + k = 14 + 2k
-2k + k = 14 – 24
-k = -10
k = 10
so total number of pendant vertices are 10
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