Level order traversal with direction change after every two levels

Given a binary tree, print the level order traversal in such a way that first two levels are printed from left to right, next two levels are printed from right to left, then next two from left to right and so on. So, the problem is to reverse the direction of level order traversal of binary tree after every two levels.

Examples:

Input: 
            1     
          /   \
        2       3
      /  \     /  \
     4    5    6    7
    / \  / \  / \  / \ 
   8  9 3   1 4  2 7  2
     /     / \    \
    16    17  18   19
Output:
1
2 3
7 6 5 4
2 7 2 4 1 3 9 8
16 17 18 19
In the above example, first two levels
are printed from left to right, next two
levels are printed from right to left,
and then last level is printed from 
left to right.

Approach:

We make use of queue and stack here. Queue is used for performing normal level order traversal. Stack is used for reversing the direction of traversal after every two levels. 

While doing normal level order traversal, first two levels nodes are printed at the time when they are popped out from the queue. For the next two levels, we instead of printing the nodes, pushed them onto the stack. When all nodes of current level are popped out, we print the nodes in the stack. In this way, we print the nodes in right to left order by making use of the stack. Now for the next two levels we again do normal level order traversal for printing nodes from left to right. Then for the next two nodes, we make use of the stack for achieving right to left order. 

In this way, we will achieve desired modified level order traversal by making use of queue and stack. 

Implementation:

C++

// CPP program to print Zig-Zag traversal
// in groups of size 2.
#include <iostream>
#include <queue>
#include <stack>
using namespace std;

// A Binary Tree Node
struct Node
{
    struct Node* left;
    int data;
    struct Node* right;
};

/* Function to print the level order of
   given binary tree. Direction of printing
   level order traversal of binary tree changes
   after every two levels */
void modifiedLevelOrder(struct Node* node)
{
    // For null root
    if (node == NULL)
        return;

    if (node->left == NULL &&
                      node->right == NULL)
    {
        cout << node->data;
        return;
    }

    // Maintain a queue for normal
    // level order traversal
    queue<Node*> myQueue;

    /* Maintain a stack for printing nodes in reverse
       order after they are popped out from queue.*/
    stack<Node*> myStack;

    struct Node* temp = NULL;

    // sz is used for storing the count
    // of nodes in a level
    int sz;

    // Used for changing the direction
    // of level order traversal
    int ct = 0;

    // Used for changing the direction
    // of level order traversal
    bool rightToLeft = false;

    // Push root node to the queue
    myQueue.push(node);

    // Run this while loop till queue got empty
    while (!myQueue.empty())
    {
        ct++;

        sz = myQueue.size();

        // Do a normal level order traversal
        for (int i = 0; i < sz; i++)
        {
            temp = myQueue.front();
            myQueue.pop();

            /*For printing nodes from left to right,
            simply print the nodes in the order in which
            they are being popped out from the queue.*/
            if (rightToLeft == false)
                cout << temp->data << " ";

            /* For printing nodes
            from right to left,
            push the nodes to stack
            instead of printing them.*/
            else
                myStack.push(temp);

            if (temp->left)
                myQueue.push(temp->left);

            if (temp->right)
                myQueue.push(temp->right);
        }

        if (rightToLeft == true)
        {

            // for printing the nodes in order
            // from right to left
            while (!myStack.empty())
            {
                temp = myStack.top();
                myStack.pop();

                cout << temp->data << " ";
            }
        }

        /*Change the direction of printing
        nodes after every two levels.*/
        if (ct == 2)
        {
            rightToLeft = !rightToLeft;
            ct = 0;
        }

        cout << "\n";
    }
}

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

// Driver program to test above functions
int main()
{
    // Let us create binary tree
    Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->left = newNode(6);
    root->right->right = newNode(7);
    root->left->left->left = newNode(8);
    root->left->left->right = newNode(9);
    root->left->right->left = newNode(3);
    root->left->right->right = newNode(1);
    root->right->left->left = newNode(4);
    root->right->left->right = newNode(2);
    root->right->right->left = newNode(7);
    root->right->right->right = newNode(2);
    root->left->right->left->left = newNode(16);
    root->left->right->left->right = newNode(17);
    root->right->left->right->left = newNode(18);
    root->right->right->left->right = newNode(19);

    modifiedLevelOrder(root);

    return 0;
}

Java

// Java program to print Zig-Zag traversal
// in groups of size 2.
import java.util.*;

public class GFG
{

// A Binary Tree Node
static class Node
{
    Node left;
    int data;
    Node right;
};

/* Function to print the level order of
given binary tree. Direction of printing
level order traversal of binary tree changes
after every two levels */
static void modifiedLevelOrder(Node node)
{
    // For null root
    if (node == null)
        return;

    if (node.left == null && node.right == null)
    {
        System.out.print(node.data);
        return;
    }

    // Maintain a queue for normal
    // level order traversal
    Queue<Node> myQueue = new LinkedList<>();

    /* Maintain a stack for
    printing nodes in reverse
    order after they are popped
    out from queue.*/
    Stack<Node> myStack = new Stack<>();

    Node temp = null;

    // sz is used for storing
    // the count of nodes in a level
    int sz;

    // Used for changing the direction
    // of level order traversal
    int ct = 0;

    // Used for changing the direction
    // of level order traversal
    boolean rightToLeft = false;

    // Push root node to the queue
    myQueue.add(node);

    // Run this while loop till queue got empty
    while (!myQueue.isEmpty())
    {
        ct++;

        sz = myQueue.size();

        // Do a normal level order traversal
        for (int i = 0; i < sz; i++)
        {
            temp = myQueue.peek();
            myQueue.remove();

            /*For printing nodes from left to right,
            simply print the nodes in the order in which
            they are being popped out from the queue.*/
            if (rightToLeft == false)
                System.out.print(temp.data + " ");

            /* For printing nodes from right to left,
            push the nodes to stack instead of printing them.*/
            else
                myStack.push(temp);

            if (temp.left != null)
                myQueue.add(temp.left);

            if (temp.right != null)
                myQueue.add(temp.right);
        }

        if (rightToLeft == true)
        {

            // for printing the nodes in order
            // from right to left
            while (!myStack.isEmpty())
            {
                temp = myStack.peek();
                myStack.pop();

                System.out.print(temp.data + " ");
            }
        }

        /*Change the direction of printing
        nodes after every two levels.*/
        if (ct == 2)
        {
            rightToLeft = !rightToLeft;
            ct = 0;
        }

        System.out.print("\n");
    }
}

// Utility function to create a new tree node
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
}

// Driver Code
public static void main(String[] args)
{
    // Let us create binary tree
    Node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.left = newNode(6);
    root.right.right = newNode(7);
    root.left.left.left = newNode(8);
    root.left.left.right = newNode(9);
    root.left.right.left = newNode(3);
    root.left.right.right = newNode(1);
    root.right.left.left = newNode(4);
    root.right.left.right = newNode(2);
    root.right.right.left = newNode(7);
    root.right.right.right = newNode(2);
    root.left.right.left.left = newNode(16);
    root.left.right.left.right = newNode(17);
    root.right.left.right.left = newNode(18);
    root.right.right.left.right = newNode(19);

    modifiedLevelOrder(root);
}
}

// This code is contributed by 29AjayKumar

Python3

# A Binary Tree Node
from collections import deque

class Node:
    def __init__(self, x):
        self.data = x
        self.left = None
        self.right = None

# /* Function to print level order of
#    given binary tree. Direction of printing
#    level order traversal of binary tree changes
#    after every two levels */
def modifiedLevelOrder(node):

    # For null root
    if (node == None):
        return

    if (node.left == None and node.right == None):
        print(node.data, end = " ")
        return

    # Maintain a queue for normal
    # level order traversal
    myQueue = deque()

    # /* Maintain a stack for printing nodes in reverse
    #    order after they are popped out from queue.*/
    myStack = []

    temp = None

    # sz is used for storing the count
    # of nodes in a level
    sz = 0

    # Used for changing the direction
    # of level order traversal
    ct = 0

    # Used for changing the direction
    # of level order traversal
    rightToLeft = False

    # Push root node to the queue
    myQueue.append(node)

    # Run this while loop till queue got empty
    while (len(myQueue) > 0):
        ct += 1

        sz = len(myQueue)

        # Do a normal level order traversal
        for i in range(sz):
            temp = myQueue.popleft()

            # /*For printing nodes from left to right,
            # simply print nodes in the order in which
            # they are being popped out from the queue.*/
            if (rightToLeft == False):
                print(temp.data,end=" ")

            # /* For printing nodes
            # from right to left,
            # push the nodes to stack
            # instead of printing them.*/
            else:
                myStack.append(temp)

            if (temp.left):
                myQueue.append(temp.left)

            if (temp.right):
                myQueue.append(temp.right)

        if (rightToLeft == True):

            # for printing the nodes in order
            # from right to left
            while (len(myStack) > 0):
                temp = myStack[-1]
                del myStack[-1]

                print(temp.data, end = " ")

        # /*Change the direction of printing
        # nodes after every two levels.*/
        if (ct == 2):
            rightToLeft = not rightToLeft
            ct = 0

        print()

# Driver program to test above functions
if __name__ == '__main__':

    # Let us create binary tree
    root = Node(1)
    root.left = Node(2)
    root.right = Node(3)
    root.left.left = Node(4)
    root.left.right = Node(5)
    root.right.left = Node(6)
    root.right.right = Node(7)
    root.left.left.left = Node(8)
    root.left.left.right = Node(9)
    root.left.right.left = Node(3)
    root.left.right.right = Node(1)
    root.right.left.left = Node(4)
    root.right.left.right = Node(2)
    root.right.right.left = Node(7)
    root.right.right.right = Node(2)
    root.left.right.left.left = Node(16)
    root.left.right.left.right = Node(17)
    root.right.left.right.left = Node(18)
    root.right.right.left.right = Node(19)

    modifiedLevelOrder(root)

    # This code is contributed by mohit kumar 29.

C#

// C# program to print Zig-Zag traversal
// in groups of size 2.
using System;
using System.Collections.Generic;
class GFG
{

// A Binary Tree Node
public class Node
{
    public Node left;
    public int data;
    public Node right;
};

/* Function to print the level order of
given binary tree. Direction of printing
level order traversal of binary tree changes
after every two levels */
static void modifiedLevelOrder(Node node)
{
    // For null root
    if (node == null)
        return;

    if (node.left == null &&
        node.right == null)
    {
        Console.Write(node.data);
        return;
    }

    // Maintain a queue for
    // normal level order traversal
    Queue<Node> myQueue = new Queue<Node>();

    /* Maintain a stack for printing nodes
    in reverse order after they are
    popped out from queue.*/
    Stack<Node> myStack = new Stack<Node>();
    Node temp = null;

    // sz is used for storing
    // the count of nodes in a level
    int sz;

    // Used for changing the direction
    // of level order traversal
    int ct = 0;

    // Used for changing the direction
    // of level order traversal
    bool rightToLeft = false;

    // Push root node to the queue
    myQueue.Enqueue(node);

    // Run this while loop
    // till queue got empty
    while (myQueue.Count != 0)
    {
        ct++;
        sz = myQueue.Count;

        // Do a normal level order traversal
        for (int i = 0; i < sz; i++)
        {
            temp = myQueue.Peek();
            myQueue.Dequeue();

            /* For printing nodes from left to right,
            simply print the nodes in the order in which
            they are being popped out from the queue.*/
            if (rightToLeft == false)
                Console.Write(temp.data + " ");

            /* For printing nodes from right to left,
               push the nodes to stack instead of
               printing them.*/
            else
                myStack.Push(temp);
            if (temp.left != null)
                myQueue.Enqueue(temp.left);
            if (temp.right != null)
                myQueue.Enqueue(temp.right);
        }
        if (rightToLeft == true)
        {

            // for printing the nodes in order
            // from right to left
            while (myStack.Count != 0)
            {
                temp = myStack.Peek();
                myStack.Pop();

                Console.Write(temp.data + " ");
            }
        }

        /*Change the direction of printing
        nodes after every two levels.*/
        if (ct == 2)
        {
            rightToLeft = !rightToLeft;
            ct = 0;
        }
        Console.Write("\n");
    }
}

// Utility function to create a new tree node
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
}

// Driver Code
public static void Main(String[] args)
{

    // Let us create binary tree
    Node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.left = newNode(6);
    root.right.right = newNode(7);
    root.left.left.left = newNode(8);
    root.left.left.right = newNode(9);
    root.left.right.left = newNode(3);
    root.left.right.right = newNode(1);
    root.right.left.left = newNode(4);
    root.right.left.right = newNode(2);
    root.right.right.left = newNode(7);
    root.right.right.right = newNode(2);
    root.left.right.left.left = newNode(16);
    root.left.right.left.right = newNode(17);
    root.right.left.right.left = newNode(18);
    root.right.right.left.right = newNode(19);

    modifiedLevelOrder(root);
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>
// Javascript program to print Zig-Zag traversal
// in groups of size 2.

// A Binary Tree Node
class Node
{
    constructor(data)
    {
        this.data = data;
        this.left = this.right=null;
    }
}

/* Function to print the level order of
given binary tree. Direction of printing
level order traversal of binary tree changes
after every two levels */
function modifiedLevelOrder(node)
{
    // For null root
    if (node == null)
        return;

    if (node.left == null && node.right == null)
    {
        document.write(node.data);
        return;
    }

    // Maintain a queue for normal
    // level order traversal
    let myQueue = [];

    /* Maintain a stack for
    printing nodes in reverse
    order after they are popped
    out from queue.*/
    let myStack = [];

    let temp = null;

    // sz is used for storing
    // the count of nodes in a level
    let sz;

    // Used for changing the direction
    // of level order traversal
    let ct = 0;

    // Used for changing the direction
    // of level order traversal
    let rightToLeft = false;

    // Push root node to the queue
    myQueue.push(node);

    // Run this while loop till queue got empty
    while (myQueue.length != 0)
    {
        ct++;

        sz = myQueue.length;

        // Do a normal level order traversal
        for (let i = 0; i < sz; i++)
        {
            temp = myQueue.shift();


            /*For printing nodes from left to right,
            simply print the nodes in the order in which
            they are being popped out from the queue.*/
            if (rightToLeft == false)
                document.write(temp.data + " ");

            /* For printing nodes from right to left,
            push the nodes to stack instead of printing them.*/
            else
                myStack.push(temp);

            if (temp.left != null)
                myQueue.push(temp.left);

            if (temp.right != null)
                myQueue.push(temp.right);
        }

        if (rightToLeft == true)
        {

            // for printing the nodes in order
            // from right to left
            while (myStack.length != 0)
            {
                temp = myStack.pop();


                document.write(temp.data + " ");
            }
        }

        /*Change the direction of printing
        nodes after every two levels.*/
        if (ct == 2)
        {
            rightToLeft = !rightToLeft;
            ct = 0;
        }

        document.write("<br>");
    }
}

// Driver Code

// Let us create binary tree
let root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.left = new Node(6);
root.right.right = new Node(7);
root.left.left.left = new Node(8);
root.left.left.right = new Node(9);
root.left.right.left = new Node(3);
root.left.right.right = new Node(1);
root.right.left.left = new Node(4);
root.right.left.right = new Node(2);
root.right.right.left = new Node(7);
root.right.right.right = new Node(2);
root.left.right.left.left = new Node(16);
root.left.right.left.right = new Node(17);
root.right.left.right.left = new Node(18);
root.right.right.left.right = new Node(19);

modifiedLevelOrder(root);

// This code is contributed by unknown2108
</script>

Output

1 
2 3 
7 6 5 4 
2 7 2 4 1 3 9 8 
16 17 18 19 

Time Complexity: O(n). Each node is traversed at most twice while doing level order traversal, so time complexity would be O(n).
Auxiliary  Space: O(n) since the maximum size of the stack or queue can grow up to the number of nodes in the binary tree.

Approach 2:

We make use of queue and stack here but in a different way. Using macros #define ChangeDirection(Dir) ((Dir) = 1 – (Dir)). In following implementation directs the order of push operations in both queue or stack. 

In this way, we will achieve desired modified level order traversal by making use of queue and stack.

Implementation:

C++

// CPP program to print Zig-Zag traversal
// in groups of size 2.
#include <iostream>
#include <stack>
#include <queue>

using namespace std;

#define LEFT 0
#define RIGHT 1
#define ChangeDirection(Dir) ((Dir) = 1 - (Dir))

// A Binary Tree Node
struct node
{
    int data;
    struct node *left, *right;
};

// Utility function to create a new tree node
node* newNode(int data)
{
    node* temp = new node;
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}

/* Function to print the level order of
   given binary tree. Direction of printing
   level order traversal of binary tree changes
   after every two levels */
void modifiedLevelOrder(struct node *root)
{
    if (!root)
        return ;

    int dir = LEFT;
    struct node *temp;
    queue <struct node *> Q;
    stack <struct node *> S;

    S.push(root);

    // Run this while loop till queue got empty
    while (!Q.empty() || !S.empty())
    {
        while (!S.empty())
        {
            temp = S.top();
            S.pop();
            cout << temp->data << " ";

            if (dir == LEFT) {
                if (temp->left)
                    Q.push(temp->left);
                if (temp->right)
                    Q.push(temp->right);
            }
            /* For printing nodes from right to left,
            push the nodes to stack
             instead of printing them.*/
            else {
                if (temp->right)
                    Q.push(temp->right);
                if (temp->left)
                    Q.push(temp->left);
            }
        }

        cout << endl;

            // for printing the nodes in order
            // from right to left
        while (!Q.empty())
        {
            temp = Q.front();
            Q.pop();
            cout << temp->data << " ";

            if (dir == LEFT) {
                if (temp->left)
                    S.push(temp->left);
                if (temp->right)
                    S.push(temp->right);
            } else {
                if (temp->right)
                    S.push(temp->right);
                if (temp->left)
                    S.push(temp->left);
            }
        }
        cout << endl;

        // Change the direction of traversal.
        ChangeDirection(dir);
    }
}

// Driver program to test above functions
int main()
{
    // Let us create binary tree
    node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->left = newNode(6);
    root->right->right = newNode(7);
    root->left->left->left = newNode(8);
    root->left->left->right = newNode(9);
    root->left->right->left = newNode(3);
    root->left->right->right = newNode(1);
    root->right->left->left = newNode(4);
    root->right->left->right = newNode(2);
    root->right->right->left = newNode(7);
    root->right->right->right = newNode(2);
    root->left->right->left->left = newNode(16);
    root->left->right->left->right = newNode(17);
    root->right->left->right->left = newNode(18);
    root->right->right->left->right = newNode(19);

    modifiedLevelOrder(root);

    return 0;
}

Java

// JAVA program to print Zig-Zag traversal
// in groups of size 2.
import java.util.*;
public class GFG
{

static final int LEFT = 0;
static final int RIGHT = 1;
static  int ChangeDirection(int Dir)
{

Dir = 1 - Dir;
return Dir;
}

// A Binary Tree Node
static class node
{
    int data;
    node left, right;
};

// Utility function to create a new tree node
static node newNode(int data)
{
    node temp = new node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
}

/* Function to print the level order of
   given binary tree. Direction of printing
   level order traversal of binary tree changes
   after every two levels */
static void modifiedLevelOrder(node root)
{
    if (root == null)
        return ;
    int dir = LEFT;
    node temp;
    Queue <node > Q = new LinkedList<>();
    Stack <node > S = new Stack<>();
    S.add(root);

    // Run this while loop till queue got empty
    while (!Q.isEmpty() || !S.isEmpty())
    {
        while (!S.isEmpty())
        {
            temp = S.peek();
            S.pop();
            System.out.print(temp.data + " ");

            if (dir == LEFT)
            {
                if (temp.left != null)
                    Q.add(temp.left);
                if (temp.right != null)
                    Q.add(temp.right);
            }

            /* For printing nodes from right to left,
            push the nodes to stack
             instead of printing them.*/
            else {
                if (temp.right != null)
                    Q.add(temp.right);
                if (temp.left != null)
                    Q.add(temp.left);
            }
        }
        System.out.println();

            // for printing the nodes in order
            // from right to left
        while (!Q.isEmpty())
        {
            temp = Q.peek();
            Q.remove();
            System.out.print(temp.data + " ");

            if (dir == LEFT) {
                if (temp.left != null)
                    S.add(temp.left);
                if (temp.right != null)
                    S.add(temp.right);
            } else {
                if (temp.right != null)
                    S.add(temp.right);
                if (temp.left != null)
                    S.add(temp.left);
            }
        }
        System.out.println();

        // Change the direction of traversal.
        dir = ChangeDirection(dir);
    }
}

// Driver code
public static void main(String[] args)
{

    // Let us create binary tree
    node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.left = newNode(6);
    root.right.right = newNode(7);
    root.left.left.left = newNode(8);
    root.left.left.right = newNode(9);
    root.left.right.left = newNode(3);
    root.left.right.right = newNode(1);
    root.right.left.left = newNode(4);
    root.right.left.right = newNode(2);
    root.right.right.left = newNode(7);
    root.right.right.right = newNode(2);
    root.left.right.left.left = newNode(16);
    root.left.right.left.right = newNode(17);
    root.right.left.right.left = newNode(18);
    root.right.right.left.right = newNode(19);
    modifiedLevelOrder(root);
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 program to print Zig-Zag traversal
# in groups of size 2.
LEFT = 0
RIGHT = 1
def ChangeDirection(Dir):
    Dir = 1 - Dir
    return Dir

# A Binary Tree Node
class node:
    # Constructor to set the data of
    # the newly created tree node
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None

""" Function to print the level order of
   given binary tree. Direction of printing
   level order traversal of binary tree changes
   after every two levels """
def modifiedLevelOrder(root):
    if (root == None):
        return
    Dir = LEFT
    Q = []
    S = []
    S.append(root)

    # Run this while loop till queue got empty
    while (len(Q) > 0 or len(S) > 0):
        while (len(S) > 0):
            temp = S[-1]
            S.pop()
            print(temp.data, end = " ")

            if (Dir == LEFT):
                if (temp.left != None):
                    Q.append(temp.left)
                if (temp.right != None):
                    Q.append(temp.right)
            else:
                if (temp.right != None):
                    Q.append(temp.right)
                if (temp.left != None):
                    Q.append(temp.left)
        print()

        # for printing the nodes in order
        # from right to left
        while len(Q) > 0:
            temp = Q[0]
            Q.pop(0)
            print(temp.data, end = " ")

            if (Dir == LEFT):
                if (temp.left != None):
                    S.append(temp.left)
                if (temp.right != None):
                    S.append(temp.right)
            else:
                if (temp.right != None):
                    S.append(temp.right)
                if (temp.left != None):
                    S.append(temp.left)
        print()

        # Change the direction of traversal.
        Dir = ChangeDirection(Dir)

# Let us create binary tree
root = node(1)
root.left = node(2)
root.right = node(3)
root.left.left = node(4)
root.left.right = node(5)
root.right.left = node(6)
root.right.right = node(7)
root.left.left.left = node(8)
root.left.left.right = node(9)
root.left.right.left = node(3)
root.left.right.right = node(1)
root.right.left.left = node(4)
root.right.left.right = node(2)
root.right.right.left = node(7)
root.right.right.right = node(2)
root.left.right.left.left = node(16)
root.left.right.left.right = node(17)
root.right.left.right.left = node(18)
root.right.right.left.right = node(19)
modifiedLevelOrder(root)

# This code is contributed by suresh07.

C#

// C# program to print Zig-Zag traversal
// in groups of size 2.
using System;
using System.Collections.Generic;


public class GFG
{

static readonly int LEFT = 0;
static readonly int RIGHT = 1;
static  int ChangeDirection(int Dir)
{

Dir = 1 - Dir;
return Dir;
}

// A Binary Tree Node
public

 class node
{
    public

 int data;
    public

 node left, right;
};

// Utility function to create a new tree node
static node newNode(int data)
{
    node temp = new node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
}

/* Function to print the level order of
   given binary tree. Direction of printing
   level order traversal of binary tree changes
   after every two levels */
static void modifiedLevelOrder(node root)
{
    if (root == null)
        return ;
    int dir = LEFT;
    node temp;
    Queue <node > Q = new Queue<node>();
    Stack <node > S = new Stack<node>();
    S.Push(root);

    // Run this while loop till queue got empty
    while (Q.Count!=0 || S.Count!=0)
    {
        while (S.Count!=0)
        {
            temp = S.Peek();
            S.Pop();
            Console.Write(temp.data + " ");

            if (dir == LEFT)
            {
                if (temp.left != null)
                    Q.Enqueue(temp.left);
                if (temp.right != null)
                    Q.Enqueue(temp.right);
            }

            /* For printing nodes from right to left,
            push the nodes to stack
             instead of printing them.*/
            else {
                if (temp.right != null)
                    Q.Enqueue(temp.right);
                if (temp.left != null)
                    Q.Enqueue(temp.left);
            }
        }
        Console.WriteLine();

            // for printing the nodes in order
            // from right to left
        while (Q.Count!=0)
        {
            temp = Q.Peek();
            Q.Dequeue();
            Console.Write(temp.data + " ");

            if (dir == LEFT) {
                if (temp.left != null)
                    S.Push(temp.left);
                if (temp.right != null)
                    S.Push(temp.right);
            } else {
                if (temp.right != null)
                    S.Push(temp.right);
                if (temp.left != null)
                    S.Push(temp.left);
            }
        }
        Console.WriteLine();

        // Change the direction of traversal.
        dir = ChangeDirection(dir);
    }
}

// Driver code
public static void Main(String[] args)
{

    // Let us create binary tree
    node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.left.right = newNode(5);
    root.right.left = newNode(6);
    root.right.right = newNode(7);
    root.left.left.left = newNode(8);
    root.left.left.right = newNode(9);
    root.left.right.left = newNode(3);
    root.left.right.right = newNode(1);
    root.right.left.left = newNode(4);
    root.right.left.right = newNode(2);
    root.right.right.left = newNode(7);
    root.right.right.right = newNode(2);
    root.left.right.left.left = newNode(16);
    root.left.right.left.right = newNode(17);
    root.right.left.right.left = newNode(18);
    root.right.right.left.right = newNode(19);
    modifiedLevelOrder(root);
}
}

// This code contributed by Rajput-Ji

Javascript

<script>
// Javascript program to print Zig-Zag traversal
// in groups of size 2.

let LEFT = 0;
let RIGHT = 1;

// A Binary Tree Node
function ChangeDirection(Dir)
{
    Dir = 1 - Dir;
return Dir;
}

class Node
{
    // Utility function to create a new tree node
    constructor(data)
    {
        this.data=data;
        this.left=this.right=null;
    }
}

/* Function to print the level order of
   given binary tree. Direction of printing
   level order traversal of binary tree changes
   after every two levels */
function modifiedLevelOrder(root)
{
    if (root == null)
        return ;
    let dir = LEFT;
    let temp;
    let Q = [];
    let S = [];
    S.push(root);

    // Run this while loop till queue got empty
    while (Q.length!=0 || S.length!=0)
    {
        while (S.length!=0)
        {
            temp = S.pop();

            document.write(temp.data + " ");

            if (dir == LEFT)
            {
                if (temp.left != null)
                    Q.push(temp.left);
                if (temp.right != null)
                    Q.push(temp.right);
            }

            /* For printing nodes from right to left,
            push the nodes to stack
             instead of printing them.*/
            else {
                if (temp.right != null)
                    Q.push(temp.right);
                if (temp.left != null)
                    Q.push(temp.left);
            }
        }
        document.write("<br>");

            // for printing the nodes in order
            // from right to left
        while (Q.length!=0)
        {
            temp = Q[0];
            Q.shift();
            document.write(temp.data + " ");

            if (dir == LEFT) {
                if (temp.left != null)
                    S.push(temp.left);
                if (temp.right != null)
                    S.push(temp.right);
            } else {
                if (temp.right != null)
                    S.push(temp.right);
                if (temp.left != null)
                    S.push(temp.left);
            }
        }
        document.write("<br>");

        // Change the direction of traversal.
        dir = ChangeDirection(dir);
    }
}

// Driver code

// Let us create binary tree
let root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.left = new Node(6);
root.right.right = new Node(7);
root.left.left.left = new Node(8);
root.left.left.right = new Node(9);
root.left.right.left = new Node(3);
root.left.right.right = new Node(1);
root.right.left.left = new Node(4);
root.right.left.right = new Node(2);
root.right.right.left = new Node(7);
root.right.right.right = new Node(2);
root.left.right.left.left = new Node(16);
root.left.right.left.right = new Node(17);
root.right.left.right.left = new Node(18);
root.right.right.left.right = new Node(19);
modifiedLevelOrder(root);



// This code is contributed by patel2127
</script>

Output

1 
2 3 
7 6 5 4 
2 7 2 4 1 3 9 8 
16 17 18 19 

Time Complexity: every node is also traversed twice. There time complexity is still O(n).
Auxiliary Space: O(n), as we need to store all the nodes in the queue and stack in order to traverse the tree.

Alternative Approach (using recursion):

We make use of recursion and an additional stack to solve the problem. This solution looks simpler.

Implementation:

C++

// CPP program to implement above approach
#include <bits/stdc++.h>
using namespace std;

// A Binary Tree Node
struct node {
    int key;
    struct node *left, *right;
};

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

vector<int> stack_struct;
void print_level(node* root,int level,bool stack);

// Function to calculate height
int heightTree(node* root)
{
    // Check if root is None
    if (root == NULL)
        return 0;
    else
    {
        int lheight = heightTree(root->left);
        int rheight = heightTree(root->right);

        // Check greater between
        // lheight and rheight
        if (lheight > rheight)
            return (1 + lheight);
        else
            return (1 + rheight);
    }
}

// Function to print 2 levels
void print_2_levels(node* root)
{
    // Check if root is None
    if(root==NULL)
        return;

    int height = heightTree(root);
    int count = 0;

    // Iterate from 1 to height
    for(int i=1;i<height+1;i++)
    {
        stack_struct = {};
        // Check is count is less than 2
        if (count < 2)
            print_level(root, i, false);
        else
        {
            print_level(root, i, true);

            // Iterate backwards from len(stack_struct)-1
            // till 0

            for(int i=stack_struct.size()-1;i>=0;i--)
              cout<< stack_struct[i]<<" ";
            if (count == 3)
                count = -1;
        }

        // Increment Counter
        count += 1;
        cout << "\n";
    }
}
// Function to print level
void print_level(node* root,int level,bool stack)
{
    // Check if root is None
    if (root==NULL)
        return;

    // Check is level is 1 and stack is False
    if (level == 1 && stack == false)
        cout<< root->key<<" ";

    // Check if level is 1 and stack is True
    else if (level == 1 && stack == true)
        stack_struct.push_back(root->key);

    // Check if level is greater than 1
    else if (level > 1)
    {
        print_level(root->left, level-1, stack);
        print_level(root->right, level-1, stack);
    }
}
// Driver program to test above functions
int main()
{
    // Let us create binary tree
    node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->left = newNode(6);
    root->right->right = newNode(7);
    root->left->left->left = newNode(8);
    root->left->left->right = newNode(9);
    root->left->right->left = newNode(3);
    root->left->right->right = newNode(1);
    root->right->left->left = newNode(4);
    root->right->left->right = newNode(2);
    root->right->right->left = newNode(7);
    root->right->right->right = newNode(2);
    root->left->right->left->left = newNode(16);
    root->left->right->left->right = newNode(17);
    root->right->left->right->left = newNode(18);
    root->right->right->left->right = newNode(19);

    cout<<"Different levels:\n";

    //Function Call
    print_2_levels(root);

    return 0;
}

// THis code is contributed by Abhijeet Kumar(abhijeet19403)

Java

// Java program to implement above approach

import java.util.*;

public class GFG {

    // A Binary Tree Node
    static class node {
        int key;
        node left, right;
    };

    // Utility function to create a new tree node
    static node newNode(int data)
    {
        node temp = new node();
        temp.key = data;
        temp.left = temp.right = null;
        return temp;
    }

    static ArrayList<Integer> stack_struct;

    // Function to calculate height
    static int heightTree(node root)
    {
        // Check if root is None
        if (root == null)
            return 0;
        else {
            int lheight = heightTree(root.left);
            int rheight = heightTree(root.right);

            // Check greater between
            // lheight and rheight
            if (lheight > rheight)
                return (1 + lheight);
            else
                return (1 + rheight);
        }
    }

    // Function to print 2 levels
    static void print_2_levels(node root)
    {
        // Check if root is None
        if (root == null)
            return;

        int height = heightTree(root);
        int count = 0;

        // Iterate from 1 to height
        for (int i = 1; i < height + 1; i++) {
            stack_struct = new ArrayList<>();
            // Check is count is less than 2
            if (count < 2)
                print_level(root, i, false);
            else {
                print_level(root, i, true);

                // Iterate backwards from
                // len(stack_struct)-1 till 0

                for (int j = stack_struct.size() - 1;
                     j >= 0; j--)
                    System.out.print(stack_struct.get(j)
                                     + " ");
                if (count == 3)
                    count = -1;
            }

            // Increment Counter
            count += 1;
            System.out.println();
        }
    }
    // Function to print level
    static void print_level(node root, int level,
                            boolean stack)
    {
        // Check if root is None
        if (root == null)
            return;

        // Check is level is 1 and stack is False
        if (level == 1 && stack == false)
            System.out.print(root.key + " ");

        // Check if level is 1 and stack is True
        else if (level == 1 && stack == true)
            stack_struct.add(root.key);

        // Check if level is greater than 1
        else if (level > 1) {
            print_level(root.left, level - 1, stack);
            print_level(root.right, level - 1, stack);
        }
    }
    // Driver program to test above functions
    public static void main(String[] args)
    {
        // Let us create binary tree
        node root = newNode(1);
        root.left = newNode(2);
        root.right = newNode(3);
        root.left.left = newNode(4);
        root.left.right = newNode(5);
        root.right.left = newNode(6);
        root.right.right = newNode(7);
        root.left.left.left = newNode(8);
        root.left.left.right = newNode(9);
        root.left.right.left = newNode(3);
        root.left.right.right = newNode(1);
        root.right.left.left = newNode(4);
        root.right.left.right = newNode(2);
        root.right.right.left = newNode(7);
        root.right.right.right = newNode(2);
        root.left.left.right.left = newNode(16);
        root.left.right.right.left = newNode(17);
        root.left.right.right.right = newNode(18);
        root.right.left.right.right = newNode(19);
        System.out.println("Different levels:");

        // Function Call
        print_2_levels(root);
    }
}

// This code is contributed by Karandeep1234

Python3

# Python program for above approach

# Node class
class Node:
    def __init__(self, key):
        self.left = None
        self.right = None
        self.key = key

# Function to calculate height
def heightTree(root):

    # Check if root is None
    if root is None:
        return 0
    else:

        lheight = heightTree(root.left)
        rheight = heightTree(root.right)

        # Check greater between
        # lheight and rheight
        if lheight > rheight:
            return 1 + lheight
        else:
            return 1 + rheight

# Function to print 2 levels
def print_2_levels(root):

    # Check if root is None
    if root is None:
        return

    height = heightTree(root)
    count = 0

    # Iterate from 1 to height
    for i in range(1, height+1):
        global stack_struct
        stack_struct = []

        # Check is count is less than 2
        if count < 2:
            print_level(root, i, stack=False)
        else:
            print_level(root, i, stack=True)

            # Iterate backwards from len(stack_struct)-1
            # till 0
            for i in range(len(stack_struct)-1, -1, -1):
                print(stack_struct[i], end=' ')
            if count == 3:
                count = -1

        # Increment Counter
        count += 1
        print("")

# Function to print level
def print_level(root, level, stack=False):
    global stack_struct

    # Check if root is None
    if root is None:
        return

    # Check is level is 1 and stack is False
    if level == 1 and stack == False:
        print(root.key, end=' ')

    # Check if level is 1 and stack is True
    elif level == 1 and stack == True:
        stack_struct.append(root.key)

    # Check if level is greater than 1
    elif level > 1:
        print_level(root.left, level-1, stack=stack)
        print_level(root.right, level-1, stack=stack)


# Driver Code
root = Node(1)
root.left      = Node(2)
root.right     = Node(3)
root.left.left  = Node(4)
root.left.right  = Node(5)
root.right.left = Node(6)
root.right.right = Node(7)
root.left.left.left = Node(8)
root.left.left.right = Node(9)
root.left.right.left = Node(3)
root.left.right.right = Node(1)
root.right.left.left = Node(4)
root.right.left.right = Node(2)
root.right.right.left = Node(7)
root.right.right.right = Node(2)
root.left.left.right.left = Node(16)
root.left.right.right.left = Node(17)
root.left.right.right.right = Node(18)
root.right.left.right.right = Node(19)
print("Different levels:")

# Function Call
print_2_levels(root)

Javascript

<script>

// JavaScript program for above approach

// Node class
class Node{
    constructor(key){
        this.left = null
        this.right = null
        this.key = key
    }
}

let stack_struct;

// Function to calculate height
function heightTree(root){

    // Check if root is null
    if(root == null)
        return 0
    else{

        let lheight = heightTree(root.left)
        let rheight = heightTree(root.right)

        // Check greater between
        // lheight and rheight
        if(lheight > rheight)
            return 1 + lheight
        else
            return 1 + rheight
    }
}

// Function to print 2 levels
function print_2_levels(root){

    // Check if root is null
    if(root == null)
        return

    let height = heightTree(root)
    let count = 0

    // Iterate from 1 to height
    for(let i=1;i<height+1;i++){

        stack_struct = []

        // Check is count is less than 2
        if(count < 2)
            print_level(root, i, stack=false)
        else{
            print_level(root, i, stack=true)

            // Iterate backwards from len(stack_struct)-1
            // till 0
            for(let i = stack_struct.length-1;i>=0;i--){
                document.write(stack_struct[i],' ')
            }
            if(count == 3)
                count = -1
        }

        // Increment Counter
        count += 1
        document.write("</br>")
    }
}

// Function to print level
function print_level(root, level, stack=false){

    // Check if root is null
    if(root==null)
        return

    // Check is level is 1 and stack is False
    if(level == 1 && stack == false)
        document.write(root.key,' ')

    // Check if level is 1 and stack is True
    else if(level == 1 && stack == true)
        stack_struct.push(root.key)

    // Check if level is greater than 1
    else if(level > 1){
        print_level(root.left, level-1, stack=stack)
        print_level(root.right, level-1, stack=stack)
    }
}


// Driver Code
let root = new Node(1)
root.left     = new Node(2)
root.right     = new Node(3)
root.left.left = new Node(4)
root.left.right = new Node(5)
root.right.left = new Node(6)
root.right.right = new Node(7)
root.left.left.left = new Node(8)
root.left.left.right = new Node(9)
root.left.right.left = new Node(3)
root.left.right.right = new Node(1)
root.right.left.left = new Node(4)
root.right.left.right = new Node(2)
root.right.right.left = new Node(7)
root.right.right.right = new Node(2)
root.left.left.right.left = new Node(16)
root.left.right.right.left = new Node(17)
root.left.right.right.right = new Node(18)
root.right.left.right.right = new Node(19)
document.write("Different levels:","<br>")

// Function Call
print_2_levels(root)

// This code is contributed by shinjanpatra

</script>

C#

using System;
using System.Collections.Generic;

public class GFG
{
    // A Binary Tree Node
    public class Node
    {
        public int key;
        public Node left, right;
    }

    // Utility function to create a new tree node
    public static Node newNode(int data)
    {
        Node temp = new Node();
        temp.key = data;
        temp.left = temp.right = null;
        return temp;
    }

    static List<int> stack_struct;

    // Function to calculate height
    public static int heightTree(Node root)
    {
        // Check if root is null
        if (root == null)
            return 0;
        else
        {
            int lheight = heightTree(root.left);
            int rheight = heightTree(root.right);

            // Check greater between
            // lheight and rheight
            if (lheight > rheight)
                return (1 + lheight);
            else
                return (1 + rheight);
        }
    }

    // Function to print 2 levels
    public static void print_2_levels(Node root)
    {
        // Check if root is null
        if (root == null)
            return;

        int height = heightTree(root);
        int count = 0;

        // Iterate from 1 to height
        for (int i = 1; i < height + 1; i++)
        {
            stack_struct = new List<int>();
            // Check if count is less than 2
            if (count < 2)
                print_level(root, i, false);
            else
            {
                print_level(root, i, true);

                // Iterate backwards from
                // stack_struct.Count-1 till 0
                for (int j = stack_struct.Count - 1;
                     j >= 0; j--)
                    Console.Write(stack_struct[j] + " ");
                if (count == 3)
                    count = -1;
            }

            // Increment Counter
            count += 1;
            Console.WriteLine();
        }
    }

    // Function to print level
    public static void print_level(Node root, int level,
                            bool stack)
    {
        // Check if root is null
        if (root == null)
            return;

        // Check if level is 1 and stack is False
        if (level == 1 && stack == false)
            Console.Write(root.key + " ");

        // Check if level is 1 and stack is True
        else if (level == 1 && stack == true)
            stack_struct.Add(root.key);

        // Check if level is greater than 1
        else if (level > 1)
        {
            print_level(root.left, level - 1, stack);
            print_level(root.right, level - 1, stack);
        }
    }

    // Driver program to test above functions
    public static void Main(string[] args)
    {
        // Let us create binary tree
        Node root = newNode(1);
        root.left = newNode(2);
        root.right = newNode(3);
        root.left.left = newNode(4);
        root.left.right = newNode(5);
        root.right.left = newNode(6);
        root.right.right = newNode(7);
        root.left.left.left = newNode(8);
        root.left.left.right = newNode(9);
        root.left.right.left = newNode(3);
        root.left.right.right = newNode(1);
        root.right.left.left = newNode(4);
        root.right.left.right = newNode(2);
        root.right.right.left = newNode(7);
        root.right.right.right = newNode(2);
        root.left.left.right.left = newNode(16);
        root.left.right.right.left = newNode(17);
        root.left.right.right.right = newNode(18);
        root.right.left.right.right = newNode(19);
        Console.WriteLine("Different levels:");
        print_2_levels(root);
    }
}

Output

Different levels:
1 
2 3 
7 6 5 4 
2 7 2 4 1 3 9 8 
16 17 18 19 

Time Complexity: O(n) where n is the number of nodes in the tree (because level order traversal is done and each element is traversed maximum twice which is 2*n times – so the order of complexity is O(n)).
Auxiliary Space: O(n) for call stack.