Level order traversal in spiral form | Using Deque

Given a Binary Tree, the task is to print spiral order traversal of the given tree. For below tree, the function should print 1, 2, 3, 4, 5, 6, 7.

Examples:

Input: 
     1
    / \
   3   2
Output :
1
3 2

Input :
        10
       /  \
      20  30
     / \  
    40 60
Output :
10
20 30
60 40

We have seen recursive and iterative solutions using two stacks and an approach using one stack and one queue. In this post, a solution with one deque is discussed. The idea is to use a direction variable and decide whether to pop elements from the front or from the rear based on the value of this direction variable.
Below is the implementation of the above approach: 

C++

// C++ program to print level order traversal
// in spiral form using one deque.
#include <bits/stdc++.h>
using namespace std;

class Node {
public:
    int data;
    Node *left, *right;
    Node(int val)
    {
        data = val;
        left = NULL;
        right = NULL;
    }
};

void spiralOrder(Node* root)
{

    deque<Node*> d;

    // Push root
    d.push_back(root);

    // Direction 0 shows print right to left
    // and for Direction 1 left to right
    int dir = 0;
    while (!d.empty()) {
        int size = d.size();
        while (size--) {
            // One whole level
            // will be print in this loop

            if (dir == 0) {
                Node* temp = d.back();
                d.pop_back();
                if (temp->right)
                    d.push_front(temp->right);
                if (temp->left)
                    d.push_front(temp->left);
                cout << temp->data << " ";
            }
            else {
                Node* temp = d.front();
                d.pop_front();
                if (temp->left)
                    d.push_back(temp->left);
                if (temp->right)
                    d.push_back(temp->right);
                cout << temp->data << " ";
            }
        }
        cout << endl;
        // Direction change
        dir = 1 - dir;
    }
}

int main()
{
    // Build the Tree
    Node* root = new Node(10);
    root->left = new Node(20);
    root->right = new Node(30);
    root->left->left = new Node(40);
    root->left->right = new Node(60);

    // Call the Function
    spiralOrder(root);

    return 0;
}

Java

// Java program to print level order traversal
// in spiral form using one deque.
import java.util.*;

class GFG
{

static class Node
{
    int data;
    Node left, right;
    Node(int val)
    {
        data = val;
        left = null;
        right = null;
    }
};

static void spiralOrder(Node root)
{

    Deque<Node> d = new LinkedList<Node>();

    // Push root
    d.addLast(root);

    // Direction 0 shows print right to left
    // and for Direction 1 left to right
    int dir = 0;
    while (d.size() > 0)
    {
        int size = d.size();
        while (size-->0)
        {
            // One whole level
            // will be print in this loop

            if (dir == 0)
            {
                Node temp = d.peekLast();
                d.pollLast();
                if (temp.right != null)
                    d.addFirst(temp.right);
                if (temp.left != null)
                    d.addFirst(temp.left);
                System.out.print(temp.data + " ");
            }
            else
            {
                Node temp = d.peekFirst();
                d.pollFirst();
                if (temp.left != null)
                    d.addLast(temp.left);
                if (temp.right != null)
                    d.addLast(temp.right);
                System.out.print(temp.data + " ");
            }
        }
        System.out.println();

        // Direction change
        dir = 1 - dir;
    }
}

// Driver code
public static void main(String args[])
{
    // Build the Tree
    Node root = new Node(10);
    root.left = new Node(20);
    root.right = new Node(30);
    root.left.left = new Node(40);
    root.left.right = new Node(60);

    // Call the Function
    spiralOrder(root);
}
}

// This code is contributed by Arnab Kundu

Python3

# Python program to print level order traversal
# in spiral form using one deque.
class Node :
    def __init__(self,val) :
        self.data = val;
        self.left = None;
        self.right = None;

def spiralOrder(root) :

    d = [];

    # Push root
    d.append(root);

    # Direction 0 shows print right to left
    # and for Direction 1 left to right
    direct = 0;
    while (len(d) != 0) :
        size = len(d);

        while (size) :
            size -= 1;

            # One whole level
            # will be print in this loop
            if (direct == 0) :
                temp = d.pop();

                if (temp.right) :
                    d.insert(0, temp.right);

                if (temp.left) :
                    d.insert(0, temp.left);

                print(temp.data, end= " ");

            else :
                temp = d[0];
                d.pop(0);

                if (temp.left) :
                    d.append(temp.left);

                if (temp.right) :
                    d.append(temp.right);


                print(temp.data ,end= " ");

        print()

        # Direction change
        direct = 1 - direct;

if __name__ == "__main__" :

    # Build the Tree
    root = Node(10);
    root.left = Node(20);
    root.right = Node(30);
    root.left.left = Node(40);
    root.left.right = Node(60);

    # Call the Function
    spiralOrder(root);

# This code is contributed by AnkitRai01

C#

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

class GFG
{
    class Node
    {
        public int data;
        public Node left, right;
        public Node(int val)
        {
            data = val;
            left = null;
            right = null;
        }
    }

    static void SpiralOrder(Node root)
    {
        Queue<Node> d = new Queue<Node>();

        // Push root
        d.Enqueue(root);

        // Direction 0 shows print right to left
        // and for Direction 1 left to right
        int dir = 0;
        while (d.Count > 0)
        {
            int size = d.Count;
            while (size-- > 0)
            {
                // One whole level
                // will be print in this loop

                if (dir == 1)
                {
                    Node temp = d.Peek();
                    d.Dequeue();
                    if (temp.right != null)
                        d.Enqueue(temp.right);
                    if (temp.left != null)
                        d.Enqueue(temp.left);
                    Console.Write(temp.data + " ");
                }
                else
                {
                    Node temp = d.Peek();
                    d.Dequeue();
                    if (temp.left != null)
                        d.Enqueue(temp.left);
                    if (temp.right != null)
                        d.Enqueue(temp.right);
                    Console.Write(temp.data + " ");
                }
            }
            Console.WriteLine();

            // Direction change
            dir = 1 - dir;
        }
    }

    // Driver code

    // Driver code
    static void Main(string[] args)
    {
        // Build the Tree
        Node root = new Node(10);
        root.left = new Node(20);
        root.right = new Node(30);
        root.left.left = new Node(40);
        root.left.right = new Node(60);

        // Call the Function
        SpiralOrder(root);
    }
}
//This code is contributed by Potta Lokesh

Javascript

<script>

// JavaScript program to print level order traversal
// in spiral form using one deque.

class Node {

    constructor(val)
    {
        this.data = val;
        this.left = null;
        this.right = null;
    }
};

function spiralOrder(root)
{

    var d = [];

    // Push root
    d.push(root);

    // Direction 0 shows print right to left
    // and for Direction 1 left to right
    var dir = 0;
    while (d.length!=0) {
        var size = d.length;
        while (size-- >0) {
            // One whole level
            // will be print in this loop

            if (dir == 0) {
                var temp = d[d.length-1];
                d.pop();
                if (temp.right!= null)
                    d.unshift(temp.right);
                if (temp.left!= null)
                    d.unshift(temp.left);
                document.write( temp.data + " ");
            }
            else {
                var temp = d[0];
                d.shift();
                if (temp.left != null)
                    d.push(temp.left);
                if (temp.right!= null)
                    d.push(temp.right);
                document.write( temp.data + " ");
            }
        }
        document.write("<br>");
        // Direction change
        dir = 1 - dir;
    }
}

// Build the Tree
var root = new Node(10);
root.left = new Node(20);
root.right = new Node(30);
root.left.left = new Node(40);
root.left.right = new Node(60);
// Call the Function
spiralOrder(root);


</script>

Output:

10 
20 30 
60 40

Time Complexity: O(N) 
Space Complexity: O(N) 
where N is the number of Nodes