Preorder Traversal of N-ary Tree Without Recursion

Given an n-ary tree, print preorder traversal of it.

Example : 
Preorder traversal of below tree is A B K N M J F D G E C H I L

The idea is to use stack like iterative preorder traversal of binary tree.

1. Create an empty stack to store nodes. 
2. Push the root node to the stack. 
3. Run a loop while the stack is not empty 
   1. Pop the top node from stack. 
   2. Print the popped node. 
   3. Store all the children of popped node onto the stack. We must store children from right to left so that leftmost node is popped first. 
4. If stack is empty then we are done.

Implementation:

C++

// C++ program to traverse an N-ary tree
// without recursion
#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;
}

// Function to traverse tree without recursion
void traverse_tree(struct Node* root)
{
    // Stack to store the nodes
    stack<Node*> nodes;

    // push the current node onto the stack
    nodes.push(root);

    // loop while the stack is not empty
    while (!nodes.empty()) {

        // store the current node and pop it from the stack
        Node* curr = nodes.top();
        nodes.pop();

        // current node has been travarsed
     if(curr != NULL)
      {
         cout << curr->key << " ";

        // store all the childrent of current node from
        // right to left.
        vector<Node*>::iterator it = curr->child.end();

        while (it != curr->child.begin()) {
            it--;
            nodes.push(*it);
        }
      }
    }
}
// Driver program
int main()
{
    /*   Let us create below tree
   *            A
   *        /  / \  \
   *       B  F   D  E
   *      / \     |  /|\
   *     K  J     G C H I
   *    / \         |   |
   *   N   M        O   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[0]->child).push_back(newNode('O'));
    (root->child[3]->child[2]->child).push_back(newNode('L'));

    traverse_tree(root);

    return 0;
}

Java

// Java program to traverse an N-ary tree
// without recursion
import java.util.ArrayList;
import java.util.Stack;

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<>();
    }
};

// Function to traverse tree without recursion
static void traverse_tree(Node root)
{

    // Stack to store the nodes
    Stack<Node> nodes = new Stack<>();

    // push the current node onto the stack
    nodes.push(root);

    // Loop while the stack is not empty
    while (!nodes.isEmpty())
    {

        // Store the current node and pop
        // it from the stack
        Node curr = nodes.pop();

        // Current node has been travarsed
        if (curr != null)
        {
            System.out.print(curr.key + " ");

            // Store all the childrent of
            // current node from right to left.
            for(int i = curr.child.size() - 1; i >= 0; i--)
            {
                nodes.add(curr.child.get(i));
            }
        }
    }
}

// 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        O   L
    */

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

    traverse_tree(root);
}
}

// This code is contributed by sanjeev2552

Python3

# Python3 program to find height of
# full binary tree
# using preorder

class newNode():

    def __init__(self, key):
        self.key = key
        # all children are stored in a list
        self.child =[]

# Function to traverse tree without recursion
def traverse_tree(root):

    # Stack to store the nodes
    nodes=[]

    # push the current node onto the stack
    nodes.append(root)

    # loop while the stack is not empty
    while (len(nodes)):

        # store the current node and pop it from the stack
        curr = nodes[0]
        nodes.pop(0)

        # current node has been travarsed
        print(curr.key,end=" ")
        # store all the childrent of current node from
        # right to left.
        for it in range(len(curr.child)-1,-1,-1):
            nodes.insert(0,curr.child[it])


# Driver program to test above functions
if __name__ == '__main__':
    """   Let us create below tree
   *            A
   *        /  / \  \
   *       B  F   D  E
   *      / \     |  /|\
   *     K  J     G C H I
   *    / \         |   |
   *   N   M        O   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[0].child).append(newNode('O'))
    (root.child[3].child[2].child).append(newNode('L'))

    traverse_tree(root)

# This code is contributed by SHUBHAMSINGH10

C#

// C# program to traverse an N-ary tree
// without recursion
using System;
using System.Collections.Generic;

public class GFG{

// Structure of a node of
// an n-ary tree
public class Node
{
    public char key;
    public List<Node> child;

    public Node(char key)
    {
        this.key = key;
        child = new List<Node>();
    }
};

// Function to traverse tree without recursion
static void traverse_tree(Node root)
{

    // Stack to store the nodes
    Stack<Node> nodes = new Stack<Node>();

    // push the current node onto the stack
    nodes.Push(root);

    // Loop while the stack is not empty
    while (nodes.Count!=0)
    {

        // Store the current node and pop
        // it from the stack
        Node curr = nodes.Pop();

        // Current node has been travarsed
        if (curr != null)
        {
            Console.Write(curr.key + " ");

            // Store all the childrent of
            // current node from right to left.
            for(int i = curr.child.Count - 1; i >= 0; i--)
            {
                nodes.Push(curr.child[i]);
            }
        }
    }
}

// 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        O   L
    */

    Node root = new Node('A');
    (root.child).Add(new Node('B'));
    (root.child).Add(new Node('F'));
    (root.child).Add(new Node('D'));
    (root.child).Add(new Node('E'));
    (root.child[0].child).Add(new Node('K'));
    (root.child[0].child).Add(new Node('J'));
    (root.child[2].child).Add(new Node('G'));
    (root.child[3].child).Add(new Node('C'));
    (root.child[3].child).Add(new Node('H'));
    (root.child[3].child).Add(new Node('I'));
    (root.child[0].child[0].child).Add(new Node('N'));
    (root.child[0].child[0].child).Add(new Node('M'));
    (root.child[3].child[0].child).Add(new Node('O'));
    (root.child[3].child[2].child).Add(new Node('L'));

    traverse_tree(root);
}
}

// This code contributed by shikhasingrajput

Javascript

<script>

// Javascript program to traverse an N-ary tree
// without recursion

// Structure of a node of
// an n-ary tree
class Node
{
    constructor(key)
    {
        this.key = key;
        this.child = [];
    }
};

// Function to traverse tree without recursion
function traverse_tree(root)
{

    // Stack to store the nodes
    var nodes = [];

    // Push the current node onto the stack
    nodes.push(root);

    // Loop while the stack is not empty
    while (nodes.length != 0)
    {

        // Store the current node and pop
        // it from the stack
        var curr = nodes.pop();

        // Current node has been travarsed
        if (curr != null)
        {
            document.write(curr.key + " ");

            // Store all the childrent of
            // current node from right to left.
            for(var i = curr.child.length - 1;
                    i >= 0; i--)
            {
                nodes.push(curr.child[i]);
            }
        }
    }
}

// Driver code
/*   Let us create below tree
*            A
*        /  / \  \
*       B  F   D  E
*      / \     |  /|\
*     K  J     G C H I
*    / \         |   |
*   N   M        O   L
*/
var root = new Node('A');
(root.child).push(new Node('B'));
(root.child).push(new Node('F'));
(root.child).push(new Node('D'));
(root.child).push(new Node('E'));
(root.child[0].child).push(new Node('K'));
(root.child[0].child).push(new Node('J'));
(root.child[2].child).push(new Node('G'));
(root.child[3].child).push(new Node('C'));
(root.child[3].child).push(new Node('H'));
(root.child[3].child).push(new Node('I'));
(root.child[0].child[0].child).push(new Node('N'));
(root.child[0].child[0].child).push(new Node('M'));
(root.child[3].child[0].child).push(new Node('O'));
(root.child[3].child[2].child).push(new Node('L'));

traverse_tree(root);

// This code is contributed by rutvik_56

</script>

Output

A B K N M J F D G E C O H I L 

Time Complexity: O(N)
Auxiliary Space: O(N)