Possible edges of a tree for given diameter, height and vertices

Find a tree with the given values and print the edges of the tree. Print “-1”, if the tree is not possible.

Given three integers n, d and h.

n -> Number of vertices. [1, n]
d -> Diameter of the tree (largest 
   distance between two vertices).
h -> Height of the tree (longest distance
   between vertex 1 and another vertex)

Examples :

Input : n = 5, d = 3, h = 2 
Output : 1 2
         2 3
         1 4
         1 5
Explanation :  

We can see that the height of the tree is 2 (1 -> 
2 --> 5) and diameter is 3 ( 3 -> 2 -> 1 -> 5).
So our conditions are satisfied.

Input :  n = 8, d = 4, h = 2
Output : 1 2
         2 3
         1 4
         4 5
         1 6
         1 7
         1 8
Explanation :

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Observe that when d = 1, we cannot construct a tree (if tree has more than 2 vertices). Also when d > 2*h, we cannot construct a tree.
As we know that height is the longest path from vertex 1 to another vertex. So build that path from vertex 1 by adding edges up to h. Now, if d > h, we should add another path to satisfy diameter from vertex 1, with a length of d – h.
Our conditions for height and diameter are satisfied. But still some vertices may be left. Add the remaining vertices at any vertex other than the end points. This step will not alter our diameter and height. Chose vertex 1 to add the remaining vertices (you can chose any).
But when d == h, choose vertex 2 to add the remaining vertices.

CPP

// CPP program to construct tree for given count 
// width and height. 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to construct the tree 
void constructTree(int n, int d, int h) 
{ 
    if (d == 1) { 
  
        // Special case when d == 2, only one edge 
        if (n == 2 && h == 1) { 
            cout << "1 2" << endl; 
            return; 
        } 
        cout << "-1" << endl; // Tree is not possible 
        return; 
    } 
  
    if (d > 2 * h) { 
        cout << "-1" << endl; 
        return; 
    } 
  
    // Satisfy the height condition by add  
    // edges up to h 
    for (int i = 1; i <= h; i++)      
        cout << i << " " << i + 1 << endl; 
      
    if (d > h) { 
  
        // Add d - h edges from 1 to  
        // satisfy diameter condition 
        cout << "1"
            << " " << h + 2 << endl; 
        for (int i = h + 2; i <= d; i++) { 
            cout << i << " " << i + 1 << endl; 
        } 
    } 
  
    // Remaining edges at vertex 1 or 2(d == h) 
    for (int i = d + 1; i < n; i++)  
    { 
        int k = 1; 
        if (d == h) 
            k = 2; 
        cout << k << " " << i + 1 << endl; 
    } 
} 
  
// Driver Code 
int main() 
{ 
    int n = 5, d = 3, h = 2; 
    constructTree(n, d, h); 
    return 0; 
} 

Java

// Java program to construct tree for given count  
// width and height.  
class GfG {  
  
// Function to construct the tree  
static void constructTree(int n, int d, int h)  
{  
    if (d == 1) {  
  
        // Special case when d == 2, only one edge  
        if (n == 2 && h == 1) {  
            System.out.println("1 2");  
            return;  
        }  
        System.out.println("-1"); // Tree is not possible  
        return;  
    }  
  
    if (d > 2 * h) {  
        System.out.println("-1");  
        return;  
    }  
  
    // Satisfy the height condition by add  
    // edges up to h  
    for (int i = 1; i <= h; i++)      
        System.out.println(i + " " + (i + 1));  
      
    if (d > h) {  
  
        // Add d - h edges from 1 to  
        // satisfy diameter condition  
        System.out.println("1" + " " + (h + 2));  
        for (int i = h + 2; i <= d; i++) {  
            System.out.println(i + " " + (i + 1));  
        }  
    }  
  
    // Remaining edges at vertex 1 or 2(d == h)  
    for (int i = d + 1; i < n; i++)  
    {  
        int k = 1;  
        if (d == h)  
            k = 2;  
        System.out.println(k + " " + (i + 1));  
    }  
}  
  
// Driver Code  
public static void main(String[] args)  
{  
    int n = 5, d = 3, h = 2;  
    constructTree(n, d, h);  
}  
}  

Python3

# Python3 code to construct tree for given count 
# width and height. 
  
# Function to construct the tree 
def constructTree(n, d, h): 
    if d == 1: 
  
        # Special case when d == 2, only one edge 
        if n == 2 and h == 1: 
            print("1 2") 
            return 0
          
        print("-1")    # Tree is not possible 
        return 0
      
    if d > 2 * h: 
        print("-1") 
        return 0
          
    # Satisfy the height condition by add  
    # edges up to h 
    for i in range(1, h+1): 
        print(i," " , i + 1) 
      
    if d > h:  
  
        # Add d - h edges from 1 to 
        # satisfy diameter condition 
        print(1,"  ", h + 2) 
        for i in range(h+2, d+1): 
            print(i, " " , i + 1) 
              
    # Remaining edges at vertex 1 or 2(d == h) 
    for i in range(d+1, n): 
        k = 1
        if d == h: 
            k = 2
        print(k ," " , i + 1) 
  
# Driver Code 
n = 5
d = 3
h = 2
constructTree(n, d, h) 
  
# This code is contributed by "Sharad_Bhardwaj". 

C#

// C# program to construct tree for   
// given count width and height.  
using System;  
  
class GfG  
{  
  
    // Function to construct the tree  
    static void constructTree(int n, int d, int h)  
    {  
        if (d == 1)  
        {  
  
            // Special case when d == 2, 
            // only one edge  
            if (n == 2 && h == 1)  
            {  
                Console.WriteLine("1 2");  
                return;  
            }  
              
            // Tree is not possible  
            Console.WriteLine("-1");  
            return;  
        }  
  
        if (d > 2 * h)  
        {  
            Console.WriteLine("-1");  
            return;  
        }  
  
        // Satisfy the height condition  
        // by add edges up to h  
        for (int i = 1; i <= h; i++)  
            Console.WriteLine(i + " " + (i + 1));  
  
        if (d > h)  
        {  
  
            // Add d - h edges from 1 to  
            // satisfy diameter condition  
            Console.WriteLine("1" + " " + (h + 2));  
            for (int i = h + 2; i <= d; i++) 
            {  
                Console.WriteLine(i + " " + (i + 1));  
            }  
        }  
  
        // Remaining edges at vertex 1 or 2(d == h)  
        for (int i = d + 1; i < n; i++)  
        {  
            int k = 1;  
            if (d == h)  
                k = 2;  
            Console.WriteLine(k + " " + (i + 1));  
        }  
    }  
  
    // Driver Code  
    public static void Main(String[] args)  
    {  
        int n = 5, d = 3, h = 2;  
        constructTree(n, d, h);  
    }  
}  
  
// This code is contributed by 29AjayKumar 

Output :

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes

Aug	SEP	Oct
	21
2019	2020	2021

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

CPP

Java

Python3

C#

Recommended Posts: