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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

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.*; 
  
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 

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 

Output:

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

Time Complexity: Each node is traversed at most twice while doing level order traversal, so time complexity would be O(n).

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.

// 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;  
} 

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).

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

C#

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