Number of ways to traverse an N-ary tree

Given an n-ary tree, count the number of ways to traverse an n-ary (or a Directed Acyclic Graph) tree starting from the root vertex

Example:

Suppose we have a given N-ary tree as shown below:

Now we have to find the number of ways of traversing the whole tree starting from the root vertex. There can be many such ways. Some of them are listed below.

• N->M->K->J->B->F->D->E->C->H->I->L->A (kind-of depth-first traversal)
• A->B->F->D->E->K->J->G->C->H->I->N->M->L (level order traversal) 

Note: The count of all ways to traverse is the product of factorials of the number of children of each node. Refer to the below figure for a clear understanding:

Here,
‘A’ has four children, so 4! permutations possible
‘B’ has two children, so 2! permutations possible
‘F’ has no children, so 0! permutations possible
….. 
And so on
Hence all such ways are- 4 ! * 2 ! * 0 ! * 1 ! * 3 ! * 2 ! * 0 ! * 0 ! * 0 ! * 0 ! * 1 ! * 0 ! * 0 ! = 576 way
That’s a huge number of ways and among them, only few prove to be useful, like- inorder, level-order, preorder, and postorder (arranged according to the popularity of these traversals)

Follow the below steps to solve the problem:

• Initialize ways equal to one and create a queue of type Node and push root into it
• While the queue is not empty
  • Initialize integer p equal to the front Node of the queue
  • Set ways equal to ways * x!, where x is the count of p’s children
  • Enqueue all the children of the Node p
• Return ways

Below is the implementation of the above approach:

C++

// C++ program to find the number of ways to traverse a
// n-ary tree starting from the root node
#include <bits/stdc++.h>
using namespace std;

// Structure of a node of an n-ary tree
struct Node {
    char key;
    vector<Node*> child;
};

// Utility function to create a new tree node
Node* newNode(int key)
{
    Node* temp = new Node;
    temp->key = key;
    return temp;
}

// Utility Function to find factorial of given number
int factorial(int n)
{
    if (n == 0)
        return 1;
    return n * factorial(n - 1);
}

// Function to calculate the number of ways of traversing
// the n-ary starting from root.
// This function is just a modified breadth-first search.
// We can use a depth-first search too.
int calculateWays(Node* root)
{
    int ways = 1; // Initialize result

    // If the tree is empty there is no way of traversing
    // the tree.
    if (root == NULL)
        return 0;

    // Create a queue and enqueue root to it.
    queue<Node*> q;
    q.push(root);

    // Level order traversal.
    while (!q.empty()) {
        // Dequeue an item from queue and print it
        Node* p = q.front();
        q.pop();

        // The number of ways is the product of
        // factorials of number of children of each node.
        ways = ways * (factorial(p->child.size()));

        // Enqueue all childrent of the dequeued item
        for (int i = 0; i < p->child.size(); i++)
            q.push(p->child[i]);
    }

    return (ways);
}

// Driver code
int main()
{
    /*   Let us create below tree
     *           A
     *         / /  \  \
     *       B  F   D  E
     *      / \     |  /|\
     *     K  J    G  C H I
     *      /\            \
     *    N   M            L
     */

    Node* root = newNode('A');
    (root->child).push_back(newNode('B'));
    (root->child).push_back(newNode('F'));
    (root->child).push_back(newNode('D'));
    (root->child).push_back(newNode('E'));
    (root->child[0]->child).push_back(newNode('K'));
    (root->child[0]->child).push_back(newNode('J'));
    (root->child[2]->child).push_back(newNode('G'));
    (root->child[3]->child).push_back(newNode('C'));
    (root->child[3]->child).push_back(newNode('H'));
    (root->child[3]->child).push_back(newNode('I'));
    (root->child[0]->child[0]->child)
        .push_back(newNode('N'));
    (root->child[0]->child[0]->child)
        .push_back(newNode('M'));
    (root->child[3]->child[2]->child)
        .push_back(newNode('L'));

    cout << calculateWays(root);
    ;

    return 0;
}

Java

//Java code for the above approach
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Queue;
public class GFG {
    // Structure of a node of an n-ary tree
    static class Node {
        char key;
        ArrayList<Node> child;
        public Node(char key)
        {
            this.key = key;
            child = new ArrayList<>();
        }
    }

    // Utility function to create a new tree node
    static Node newNode(char key) { return new Node(key); }

    // Utility Function to find factorial of given number
    static int factorial(int n)
    {
        if (n == 0) {
            return 1;
        }
        return n * factorial(n - 1);
    }

    // Function to calculate the number of ways of
    // traversing the n-ary starting from root. This
    // function is just a modified breadth-first search. We
    // can use a depth-first search too.
    static int calculateWays(Node root)
    {
        int ways = 1; // Initialize result

        // If the tree is empty there is no way of
        // traversing the tree.
        if (root == null) {
            return 0;
        }

        // Create a queue and enqueue root to it.
        Queue<Node> q = new LinkedList<>();
        q.add(root);

        // Level order traversal.
        while (!q.isEmpty()) {
            // Dequeue an item from queue and print it
            Node p = q.poll();

            // The number of ways is the product of
            // factorials of number of children of each
            // node.
            ways = ways * (factorial(p.child.size()));

            // Enqueue all children of the dequeued item
            for (int i = 0; i < p.child.size(); i++) {
                q.add(p.child.get(i));
            }
        }

        return (ways);
    }

    // Driver code
    public static void main(String[] args)
    {
        /*   Let us create below tree
         *           A
         *         / /  \  \
         *       B  F   D  E
         *      / \     |  /|\
         *     K  J    G  C H I
         *      /\            \
         *    N   M            L
         */

        Node root = newNode('A');
        root.child.add(newNode('B'));
        root.child.add(newNode('F'));
        root.child.add(newNode('D'));
        root.child.add(newNode('E'));
        root.child.get(0).child.add(newNode('K'));
        root.child.get(0).child.add(newNode('J'));
        root.child.get(2).child.add(newNode('G'));
        root.child.get(3).child.add(newNode('C'));
        root.child.get(3).child.add(newNode('H'));
        root.child.get(3).child.add(newNode('I'));
        root.child.get(0).child.get(0).child.add(
            newNode('N'));
        root.child.get(0).child.get(0).child.add(
            newNode('M'));
        root.child.get(3).child.get(2).child.add(
            newNode('L'));

        System.out.println(calculateWays(root));
    }
}

Python3

# Python3 program to find the
# number of ways to traverse a
# n-ary tree starting from the root node


class Node:
    def __init__(self, key):
        self.child = []
        self.key = key

# Utility function to create a new tree node


def newNode(key):
    temp = Node(key)
    return temp

# Utility Function to find factorial of given number


def factorial(n):

    if (n == 0):
        return 1

    return n*factorial(n-1)

# Function to calculate the number of ways of traversing
# the n-ary starting from root.
# self function is just a modified breadth-first search.
# We can use a depth-first search too.


def calculateWays(root):

    ways = 1  # Initialize result

    # If the tree is empty there is no way of traversing
    # the tree.
    if (root == None):
        return 0

    # Create a queue and enqueue root to it.
    q = []
    q.append(root)

    # Level order traversal.
    while (len(q) > 0):
        # Dequeue an item from queue and print it
        p = q[0]
        q = q[1:]

        # The number of ways is the product of
        # factorials of number of children of each node.
        ways = ways*(factorial(len(p.child)))

        # Enqueue all childrent of the dequeued item
        for i in range(len(p.child)):
            q.append(p.child[i])

    return(ways)

#  Let us create below tree
# *         A
# *         / / \ \
# *     B F D E
# *     / \     | /|\
# *     K J G C H I
# *     /\         \
# * N M         L


root = newNode('A')
(root.child).append(newNode('B'))
(root.child).append(newNode('F'))
(root.child).append(newNode('D'))
(root.child).append(newNode('E'))
(root.child[0].child).append(newNode('K'))
(root.child[0].child).append(newNode('J'))
(root.child[2].child).append(newNode('G'))
(root.child[3].child).append(newNode('C'))
(root.child[3].child).append(newNode('H'))
(root.child[3].child).append(newNode('I'))
(root.child[0].child[0].child).append(newNode('N'))
(root.child[0].child[0].child).append(newNode('M'))
(root.child[3].child[2].child).append(newNode('L'))

print(calculateWays(root))

C#

//C# code for the above approach
using System;
using System.Collections.Generic;

class Node
{
    public char Key { get; set; }
    public List<Node> Children { get; set; }

    public Node(char key)
    {
        Key = key;
        Children = new List<Node>();
    }
}

class Program
{
     // Utility Function to find factorial of given number
    static int Factorial(int n)
    {
        if (n == 0)
        {
            return 1;
        }

        return n * Factorial(n - 1);
    }
    // Function to calculate the number of ways of
    // traversing the n-ary starting from root. This
    // function is just a modified breadth-first search. We
    // can use a depth-first search too.
    static int CalculateWays(Node root)
    {
        int ways = 1;  // Initialize result

        // If the tree is empty there is no way of
        // traversing the tree.
        if (root == null)
        {
            return 0;
        }

        Queue<Node> queue = new Queue<Node>();
        queue.Enqueue(root);

        while (queue.Count != 0)
        {
            Node current = queue.Dequeue();
            ways *= Factorial(current.Children.Count);

            foreach (Node child in current.Children)
            {
                queue.Enqueue(child);
            }
        }

        return ways;
    }
    // Driver code
    static void Main(string[] args)
    {
        Node root = new Node('A');
        root.Children.Add(new Node('B'));
        root.Children.Add(new Node('F'));
        root.Children.Add(new Node('D'));
        root.Children.Add(new Node('E'));
        root.Children[0].Children.Add(new Node('K'));
        root.Children[0].Children.Add(new Node('J'));
        root.Children[2].Children.Add(new Node('G'));
        root.Children[3].Children.Add(new Node('C'));
        root.Children[3].Children.Add(new Node('H'));
        root.Children[3].Children.Add(new Node('I'));
        root.Children[0].Children[0].Children.Add(new Node('N'));
        root.Children[0].Children[0].Children.Add(new Node('M'));
        root.Children[3].Children[2].Children.Add(new Node('L'));

        Console.WriteLine(CalculateWays(root));
    }
}

Javascript

<script>

    // JavaScript program to find the
    // number of ways to traverse a
    // n-ary tree starting from the root node

    class Node
    {
        constructor(key) {
           this.child = [];
           this.key = key;
        }
    }

    // Utility function to create a new tree node
    function newNode(key)
    {
       let temp = new Node(key);
       return temp;
    }

    // Utility Function to find factorial of given number
    function factorial(n)
    {
       if (n == 0)
         return 1;
       return n*factorial(n-1);
    }

    // Function to calculate the number of ways of traversing
    // the n-ary starting from root.
    // This function is just a modified breadth-first search.
    // We can use a depth-first search too.
    function calculateWays(root)
    {
       let ways = 1;  // Initialize result

       // If the tree is empty there is no way of traversing
       // the tree.
       if (root == null)
          return 0;

       // Create a queue and enqueue root to it.
       let q = [];
       q.push(root);

       // Level order traversal.
       while (q.length > 0)
       {
             // Dequeue an item from queue and print it
             let p = q[0];
             q.shift();

             // The number of ways is the product of
             // factorials of number of children of each node.
             ways = ways*(factorial(p.child.length));

             // Enqueue all childrent of the dequeued item
             for (let i=0; i< p.child.length; i++)
                q.push(p.child[i]);
        }

       return(ways);
    }

    /*   Let us create below tree
   *           A
   *         / /  \  \
   *       B  F   D  E
   *      / \     |  /|\
   *     K  J    G  C H I
   *      /\            \
   *    N   M            L
   */

   let root = newNode('A');
   (root.child).push(newNode('B'));
   (root.child).push(newNode('F'));
   (root.child).push(newNode('D'));
   (root.child).push(newNode('E'));
   (root.child[0].child).push(newNode('K'));
   (root.child[0].child).push(newNode('J'));
   (root.child[2].child).push(newNode('G'));
   (root.child[3].child).push(newNode('C'));
   (root.child[3].child).push(newNode('H'));
   (root.child[3].child).push(newNode('I'));
   (root.child[0].child[0].child).push(newNode('N'));
   (root.child[0].child[0].child).push(newNode('M'));
   (root.child[3].child[2].child).push(newNode('L'));

   document.write(calculateWays(root));

</script>

Output

576

Time Complexity: O(N²). We can optimize the solution to work in O(N) time by per-computing factorials of all numbers from 1 to N
Auxiliary Space: O(N)

Common Pitfalls:
Since products of factorials can tend to grow very huge, so it may overflow. It is preferable to use data types like- unsigned long long int in C/C++, as the number of ways can never be a negative number. In Java and Python, there are Big Integers to take care of overflows

This article is contributed by Rachit Belwariar. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.