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.

Image

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++

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// 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;
}

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Java














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// 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 



















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Python3














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# 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 



















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Output:


10 
20 30 
60 40





Time Complexity: O(N)

Space Complexity: O(N)

where N is the number of Nodes


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