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📜  确定N元树中的Leaf节点数

📅  最后修改于: 2021-06-01 00:49:28             🧑  作者: Mango

给定‘N’‘I’的值。这里, I  表示N元树中存在的内部节点的数量,并且N元的每个节点都可以具有N  个孩子或零个孩子。任务是确定n元树中的Leaf节点数

例子

公式:

派生:该树是N元树。假设它有T个总节点,这是内部节点(I)和叶节点(L)的总和。总节点数为T的树将具有(T – 1)个边或分支。
换句话说,由于树是N元树,因此每个内部节点将具有N个分支,这些分支总共贡献了N * I个内部分支。因此,根据以上说明,我们具有以下关系:

  • N * I = T – 1
  • L + I = T

从以上两个方程式,我们可以说L =(N – 1)* I + 1。

下面是上述方法的实现:

C++
// CPP program to find number
// of leaf nodes
 
#include 
using namespace std;
 
// Function to calculate
// leaf nodes in n-ary tree
int calcNodes(int N, int I)
{
    int result = 0;
 
    result = I * (N - 1) + 1;
 
    return result;
}
 
// Driver code
int main()
{
    int N = 5, I = 2;
 
    cout << "Leaf nodes = " << calcNodes(N, I);
 
    return 0;
}


Java
// Java program to find number
// of leaf nodes
 
class GfG
{
 
// Function to calculate
// leaf nodes in n-ary tree
static int calcNodes(int N, int I)
{
    int result = 0;
 
    result = I * (N - 1) + 1;
 
    return result;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 5, I = 2;
 
    System.out.println("Leaf nodes = " +
                        calcNodes(N, I));
}
}
 
// This code is contributed by Prerna Saini


Python3
# Python3 program to find number
# of leaf nodes
 
# Function to calculate
# leaf nodes in n-ary tree
def calcNodes(N, I):
    result = 0
 
    result = I * (N - 1) + 1
 
    return result
 
# Driver Code
if __name__ == '__main__':
    N = 5
    I = 2
 
    print("Leaf nodes = ",
           calcNodes(N, I))
 
# This code is contributed
# by SHUBHAMSINGH10


C#
// C# program to find number
// of leaf nodes
using System;
 
class GFG
{
 
// Function to calculate
// leaf nodes in n-ary tree
static int calcNodes(int N, int I)
{
    int result = 0;
 
    result = I * (N - 1) + 1;
 
    return result;
}
 
// Driver code
public static void Main()
{
    int N = 5, I = 2;
 
    Console.Write("Leaf nodes = " +
                  calcNodes(N, I));
}
}
 
// This code is contributed
// by Akanksha Rai


PHP


Javascript


输出:
Leaf nodes = 9