Minimum number of nodes in an AVL Tree with given height

Given the height of an AVL tree ‘h’, the task is to find the minimum number of nodes the tree can have.

Examples :

Input : H = 0
Output : N = 1
Only '1' node is possible if the height 
of the tree is '0' which is the root node.

Input : H = 3
Output : N = 7

Recursive Approach : In an AVL tree, we have to maintain the height balance property, i.e. difference in the height of the left and the right subtrees can not be other than -1, 0 or 1 for each node.
We will try to create a recurrence relation to find minimum number of nodes for a given height, n(h).

  • For height = 0, we can only have a single node in an AVL tree, i.e. n(0) = 1
  • For height = 1, we can have a minimum of two nodes in an AVL tree, i.e. n(1) = 2
  • Now for any height ‘h’, root will have two subtrees (left and right). Out of which one has to be of height h-1 and other of h-2. [root node excluded]
  • So, n(h) = 1 + n(h-1) + n(h-2) is the required recurrence relation for h>=2 [1 is added for the root node]

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find
// minimum number of nodes
int AVLnodes(int height)
{
    // Base Conditions
    if (height == 0)
        return 1;
    else if (height == 1)
        return 2;
  
    // Recursive function call
    // for the recurrence relation
    return (1 + AVLnodes(height - 1) + AVLnodes(height - 2));
}
  
// Driver Code
int main()
{
    int H = 3;
    cout << AVLnodes(H) << endl;
}

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Java

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// Java implementation of the approach
  
class GFG{
      
  
// Function to find
// minimum number of nodes
static int AVLnodes(int height)
{
    // Base Conditions
    if (height == 0)
        return 1;
    else if (height == 1)
        return 2;
   
    // Recursive function call
    // for the recurrence relation
    return (1 + AVLnodes(height - 1) + AVLnodes(height - 2));
}
   
// Driver Code
public static void main(String args[])
{
    int H = 3;
    System.out.println(AVLnodes(H));
}
}

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Python3

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# Python 3 implementation of the approach
  
# Function to find minimum 
# number of nodes
def AVLnodes(height):
      
    # Base Conditions
    if (height == 0):
        return 1
    elif (height == 1):
        return 2
  
    # Recursive function call
    # for the recurrence relation
    return (1 + AVLnodes(height - 1) + 
                AVLnodes(height - 2))
  
# Driver Code
if __name__ == '__main__':
    H = 3
    print(AVLnodes(H))
      
# This code is contributed by
# Surendra_Gangwar

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
      
// Function to find
// minimum number of nodes
static int AVLnodes(int height)
{
    // Base Conditions
    if (height == 0)
        return 1;
    else if (height == 1)
        return 2;
  
    // Recursive function call
    // for the recurrence relation
    return (1 + AVLnodes(height - 1) + 
                AVLnodes(height - 2));
}
  
// Driver Code
public static void Main()
{
    int H = 3;
    Console.Write(AVLnodes(H));
}
}
  
// This code is contributed 
// by Akanksha Rai

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PHP

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<?php
// PHP implementation of the approach
  
// Function to find minimum
// number of nodes
function AVLnodes($height)
{
    // Base Conditions
    if ($height == 0)
        return 1;
    else if ($height == 1)
        return 2;
  
    // Recursive function call
    // for the recurrence relation
    return (1 + AVLnodes($height - 1) + 
                AVLnodes($height - 2));
}
  
// Driver Code
$H = 3;
echo(AVLnodes($H));
  
// This code is contributed
// by Code_Mech.

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Output:

7

Tail Recursive Approach :

  • The recursive function for finding n(h) (minimum number of nodes possible in an AVL Tree with height ‘h’) is n(h) = 1 + n(h-1) + n(h-2) ; h>=2 ; n(0)=1 ; n(1)=2;
  • To create a Tail Recursive Function, we will maintain 1 + n(h-1) + n(h-2) as function arguments such that rather than calculating it, we directly return its value to main function.

Below is the implementation of the above approach :

C++

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// C++ implementation of the approach
#include <iostream>
using namespace std;
  
// Function to return 
//minimum number of nodes
int AVLtree(int H, int a = 1, int b = 2)
{
    // Base Conditions
    if (H == 0)
        return 1;
    if (H == 1)
        return b;
  
    // Tail Recursive Call
    return AVLtree(H - 1, b, a + b + 1);
}
  
// Driver Code
int main()
{
    int H = 5;
    int answer = AVLtree(H);
  
    // Output the result
    cout << "n(" << H << ") = "
         << answer << endl;
    return 0;
}

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Java

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// Java implementation of the approach
class GFG
{
  
// Function to return 
//minimum number of nodes
static int AVLtree(int H, int a, int b)
{
    // Base Conditions
    if (H == 0)
        return 1;
    if (H == 1)
        return b;
  
    // Tail Recursive Call
    return AVLtree(H - 1, b, a + b + 1);
}
  
// Driver Code
public static void main(String[] args)
{
    int H = 5;
    int answer = AVLtree(H, 1, 2);
  
    // Output the result
    System.out.println("n(" + H + ") = " + answer);
}
}
  
// This code is contributed by PrinciRaj1992

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Python3

# Python3 implementation of the approach

# Function to return
# minimum number of nodes
def AVLtree(H, a, b):

# Base Conditions
if(H == 0):
return 1;
if(H == 1):
return b;

# Tail Recursive Call
return AVLtree(H – 1, b, a + b + 1);

# Driver Code
if __name__ == ‘__main__’:
H = 5;
answer = AVLtree(H, 1, 2);

# Output the result
print(“n(“, H , “) = “\
, answer);

# This code is contributed by 29AjayKumar

C#

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// C# implementation of the approach
using System;
      
class GFG
{
  
// Function to return 
//minimum number of nodes
static int AVLtree(int H, int a, int b)
{
    // Base Conditions
    if (H == 0)
        return 1;
    if (H == 1)
        return b;
  
    // Tail Recursive Call
    return AVLtree(H - 1, b, a + b + 1);
}
  
// Driver Code
public static void Main(String[] args)
{
    int H = 5;
    int answer = AVLtree(H, 1, 2);
  
    // Output the result
    Console.WriteLine("n(" + H + ") = " + answer);
}
}
  
// This code is contributed by Princi Singh

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Output:

n(5) = 20


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