Prufer Code to Tree Creation
Last Updated :
10 Mar, 2023
What is Prufer Code?
Given a tree (represented as graph, not as a rooted tree) with n labeled nodes with labels from 1 to n, a Prufer code uniquely identifies the tree. The sequence has n-2 values.
How to get Prufer Code of a tree?
- Initialize Prufer code as empty.
- Start with a leaf of lowest label say x. Find the vertex connecting it to the rest of tree say y. Remove x from the tree and add y to the Prufer Code
- Repeat above step 2 until we are left with two nodes.
A tree with labels from 1 to n.
5
/ \
1 4
/ \
2 3
PruferCode = {}
The lowest label leaf is 2, we remove it from tree
and add the other vertex (connecting it to the tree)
to Prufer code
Tree now becomes
5
/ \
1 4
\
3
Prufer Code becomes = {1}
The lowest label leaf is 3, we remove it from tree
and add the other vertex (connecting it to the tree)
to Prufer code
Tree now becomes
5
/ \
1 4
Prufer Code becomes = {1, 1}
The lowest label leaf is 1, we remove it from tree
and add the other vertex (connecting it to the tree)
to Prufer code
Tree now becomes
5
\
4
Prufer Code becomes = {1, 1, 5}
We have only two nodes left now, so we stop.
How to construct a tree from given Prufer Code?
Input : (4, 1, 3, 4)
Output : Edges of following tree
2----4----3----1----5
|
6
Input : (1, 3, 5)
Output : Edges of following tree
2----1----3----5----4
Let the length of given Prufer code be m. The idea is to create an empty graph of m+2 vertices. We remove first element from sequence. Let first element of current sequence be x. Then we find the least value which is not present in the given sequence and not yet added to the tree. Let this value be y. We add an edge from x to y and repeat this step.
Let us understand algorithm to construct tree with above first example:
Input : (4, 1, 3, 4)
Step 1: First we create an empty graph of 6 vertices
and get 4 from the sequence.
Step 2: Out of 1 to 6, the least vertex not in
Prufer sequence is 2.
Step 3: We form an edge between 2 and 4.
2----4 1 3 5 6
Step 4: Next in the sequence is 1 and corresponding
vertex with least degree is 5 (as 2 has been
considered).
2----4 1----5 3 6
Step 5: Next in the sequence is 3 and corresponding
vertex with least degree is 1
(as 1 is now not part of remaining Prufer sequence)
2----4 3----1----5 6
Step 6: Next in the sequence is 4 and corresponding vertex
with least degree is 3 (as 3 has not been considered
as is not present further in sequence)
2----4----3----1----5 6
Step 7: Finally two vertices are left out from 1 to 6 (4
and 6) so we join them.
2----4----3----1----5
|
6
This is the required tree on 6 vertices.
Following is the implementation.
C++
#include <bits/stdc++.h>
using namespace std;
void printTreeEdges( int prufer[], int m)
{
int vertices = m + 2;
int vertex_set[vertices];
for ( int i = 0; i < vertices; i++)
vertex_set[i] = 0;
for ( int i = 0; i < vertices - 2; i++)
vertex_set[prufer[i] - 1] += 1;
cout << "\nThe edge set E(G) is :\n" ;
int j = 0;
for ( int i = 0; i < vertices - 2; i++) {
for (j = 0; j < vertices; j++) {
if (vertex_set[j] == 0) {
vertex_set[j] = -1;
cout << "(" << (j + 1) << ", "
<< prufer[i] << ") " ;
vertex_set[prufer[i] - 1]--;
break ;
}
}
}
j = 0;
for ( int i = 0; i < vertices; i++) {
if (vertex_set[i] == 0 && j == 0) {
cout << "(" << (i + 1) << ", " ;
j++;
}
else if (vertex_set[i] == 0 && j == 1)
cout << (i + 1) << ")\n" ;
}
}
int main()
{
int prufer[] = { 4, 1, 3, 4 };
int n = sizeof (prufer) / sizeof (prufer[0]);
printTreeEdges(prufer, n);
return 0;
}
|
Java
class GFG {
static void printTreeEdges( int prufer[], int m)
{
int vertices = m + 2 ;
int vertex_set[] = new int [vertices];
for ( int i = 0 ; i < vertices; i++)
vertex_set[i] = 0 ;
for ( int i = 0 ; i < vertices - 2 ; i++)
vertex_set[prufer[i] - 1 ] += 1 ;
System.out.print( "\nThe edge set E(G) is :\n" );
int j = 0 ;
for ( int i = 0 ; i < vertices - 2 ; i++) {
for (j = 0 ; j < vertices; j++) {
if (vertex_set[j] == 0 ) {
vertex_set[j] = - 1 ;
System.out.print( "(" + (j + 1 ) + ", "
+ prufer[i] + ") " );
vertex_set[prufer[i] - 1 ]--;
break ;
}
}
}
j = 0 ;
for ( int i = 0 ; i < vertices; i++) {
if (vertex_set[i] == 0 && j == 0 ) {
System.out.print( "(" + (i + 1 ) + ", " );
j++;
}
else if (vertex_set[i] == 0 && j == 1 )
System.out.print((i + 1 ) + ")\n" );
}
}
public static void main(String args[])
{
int prufer[] = { 4 , 1 , 3 , 4 };
int n = prufer.length;
printTreeEdges(prufer, n);
}
}
|
Python3
def printTreeEdges(prufer, m):
vertices = m + 2
vertex_set = [ 0 ] * vertices
for i in range (vertices - 2 ):
vertex_set[prufer[i] - 1 ] + = 1
print ( "The edge set E(G) is :" )
j = 0
for i in range (vertices - 2 ):
for j in range (vertices):
if (vertex_set[j] = = 0 ):
vertex_set[j] = - 1
print ( "(" , (j + 1 ), ", " ,prufer[i], ") " ,sep = " ",end = " ")
vertex_set[prufer[i] - 1 ] - = 1
break
j = 0
for i in range (vertices):
if (vertex_set[i] = = 0 and j = = 0 ):
print ( "(" , (i + 1 ), ", " , sep = " ", end=" ")
j + = 1
else if (vertex_set[i] = = 0 and j = = 1 ):
print ((i + 1 ), ")" )
prufer = [ 4 , 1 , 3 , 4 ]
n = len (prufer)
printTreeEdges(prufer, n)
|
C#
using System;
class GFG {
static void printTreeEdges( int [] prufer, int m)
{
int vertices = m + 2;
int [] vertex_set = new int [vertices];
for ( int i = 0; i < vertices; i++)
vertex_set[i] = 0;
for ( int i = 0; i < vertices - 2; i++)
vertex_set[prufer[i] - 1] += 1;
Console.Write( "\nThe edge set E(G) is :\n" );
int j = 0;
for ( int i = 0; i < vertices - 2; i++) {
for (j = 0; j < vertices; j++) {
if (vertex_set[j] == 0) {
vertex_set[j] = -1;
Console.Write( "(" + (j + 1) + ", "
+ prufer[i] + ") " );
vertex_set[prufer[i] - 1]--;
break ;
}
}
}
j = 0;
for ( int i = 0; i < vertices; i++) {
if (vertex_set[i] == 0 && j == 0) {
Console.Write( "(" + (i + 1) + ", " );
j++;
}
else if (vertex_set[i] == 0 && j == 1)
Console.Write((i + 1) + ")\n" );
}
}
public static void Main(String[] args)
{
int [] prufer = { 4, 1, 3, 4 };
int n = prufer.Length;
printTreeEdges(prufer, n);
}
}
|
Javascript
<script>
function printTreeEdges(prufer,m)
{
let vertices = m + 2;
let vertex_set = new Array(vertices);
for (let i = 0; i < vertices; i++)
vertex_set[i] = 0;
for (let i = 0; i < vertices - 2; i++)
vertex_set[prufer[i] - 1] += 1;
document.write( "<br>The edge set E(G) is :<br>" );
let j = 0;
for (let i = 0; i < vertices - 2; i++) {
for (j = 0; j < vertices; j++) {
if (vertex_set[j] == 0) {
vertex_set[j] = -1;
document.write( "(" + (j + 1) + ", "
+ prufer[i] + ") " );
vertex_set[prufer[i] - 1]--;
break ;
}
}
}
j = 0;
for (let i = 0; i < vertices; i++) {
if (vertex_set[i] == 0 && j == 0) {
document.write( "(" + (i + 1) + ", " );
j++;
}
else if (vertex_set[i] == 0 && j == 1)
document.write((i + 1) + ")\n" );
}
}
let prufer=[4, 1, 3, 4];
let n = prufer.length;
printTreeEdges(prufer, n);
</script>
|
Output
The edge set E(G) is :
(2, 4) (5, 1) (1, 3) (3, 4) (4, 6)
Time Complexity: O(n2) where n is the number of vertices in the tree.
Auxiliary Space: O(n)
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