具有给定高度的AVL树中的最小节点数

给定AVL树的高度“ h”，任务是找到该树可以具有的最小节点数。

例子 ：

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

递归方法：在AVL树中，我们必须保持高度平衡属性，即，每个子树的左右子树的高度差不能为-1、0或1。
我们将尝试创建一个递归关系，以找到给定高度n(h)的最小节点数。

• 对于height = 0，我们在AVL树中只能有一个节点，即n(0)= 1
• 对于height = 1，我们在AVL树中至少可以有两个节点，即n(1)= 2
• 现在，对于任何高度“ h”，根都将具有两个子树(左和右)。其中一个必须具有高度h-1，另一个必须具有h-2。 [排除根节点]
• 因此， n(h)= 1 + n(h-1)+ n(h-2)是h> = 2所需的递归关系[为根节点添加1]

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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;
}

Java

// 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));
}
}

Python3

# 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

C#

// 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

---

输出：

7

尾递归方法：

• 查找n(h)(在高度为“ h”的AVL树中可能的最小节点数)的递归函数为n(h)= 1 + n(h-1)+ n(h-2)； h> = 2; n(0)= 1; n(1)= 2;
• 要创建尾递归函数，我们将1 + n(h-1)+ n(h-2)保留为函数参数，这样我们无需将其值直接返回给main函数即可对其进行计算而不是对其进行计算。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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;
}

Java

// 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

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＃

// 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

输出：

n(5) = 20