Given the value of ‘N’ and ‘I’. Here, represents the number of internal nodes present in an N-ary tree and every node of the N-ary can either have childs or zero child. The task is to determine the number of Leaf nodes in n-ary tree.
Input : N = 3, I = 5
Output : Leaf nodes = 11
Input : N = 10, I = 10
Output : leaf nodes = 91
I = Number of Internal nodes.
L = Leaf Nodes.
and, N = Number of children each node can have.
Derivation: The tree is an N-ary tree. Assume it has T total nodes, which is the sum of internal nodes (I) and leaf nodes (L). A tree with T total nodes will have (T – 1) edges or branches.
In other words, since the tree is an N-ary tree, each internal node will have N branches contributing a total of N*I internal branches. Therefore we have the following relations from the above explanations,
- N * I = T – 1
- L + I = T
From the above two equations, we can say that L = (N – 1) * I + 1.
Below is the implementation of the above approach:
// C# program to find number
// of leaf nodes
// Function to calculate
// leaf nodes in n-ary tree
static int calcNodes(int N, int I)
int result = 0;
result = I * (N – 1) + 1;
// Driver code
public static void Main()
int N = 5, I = 2;
Console.Write(“Leaf nodes = ” +
// This code is contributed
// by Akanksha Rai
Leaf nodes = 9
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