Zig Zag Level order traversal of a tree using single queue

Write a function to print spiral order traversal of a tree. For below tree, function should print 1, 2, 3, 4, 5, 6, 7.

spiral_order

We have discussed naive approach and two stack based approach in Level Order with recursion and multiple stacks

The idea behind this approach is first we have to take a queue, a direction flag and a separation flag which is NULL

  1. Insert the root element into the queue and again insert NULL into the queue.
  2. For every element in the queue insert its child nodes.
  3. If a NULL is encountered then check the direction to traverse the particular level is left to right or right to left. If it’s an even level then traverse from left to right otherwise traverse the tree in right to level order i.e., from the front to the previous front i.e., from the current NULL to to the last NULL that has been visited. This continues till the last level then there the loop breaks and we print what is left (that has not printed) by checking the direction to print.

Following is the implementation of the explanation

C++

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// C++ program to print level order traversal
// in spiral form using a single dequeue
#include <bits/stdc++.h>
  
struct Node {
    int data;
    struct Node *left, *right;
};
  
// A utility function to create a new node
struct Node* newNode(int data)
{
    struct Node* node = new struct Node;
    node->data = data;
    node->left = node->right = NULL;
    return (node);
}
  
// function to print the level order traversal
void levelOrder(struct Node* root, int n)
{
    // We can just take the size as H+N which
    // implies the height of the tree with the 
    // size of the tree
    struct Node* queue[2 * n];
    int top = -1;
    int front = 1;
    queue[++top] = NULL;
    queue[++top] = root;
    queue[++top] = NULL;
  
    // struct Node* t=root;
    int prevFront = 0, count = 1;
    while (1) {
  
        struct Node* curr = queue[front];
  
        // A level separator found
        if (curr == NULL) {
  
            // If this is the only item in dequeue
            if (front == top)
                break;
  
            // Else print contents of previous level
            // according to count
            else {
                if (count % 2 == 0) {
                    for (int i = prevFront + 1; i < front; i++)
                        printf("%d ", queue[i]->data);
                }
                else {
                    for (int i = front - 1; i > prevFront; i--)
                        printf("%d ", queue[i]->data);
                }
  
                prevFront = front;
                count++;
                front++;
  
                // Insert a new level separator
                queue[++top] = NULL;
  
                continue;
            }
        }
  
        if (curr->left != NULL)
            queue[++top] = curr->left;
        if (curr->right != NULL)
            queue[++top] = curr->right;
        front++;
    }
  
    if (count % 2 == 0) {
        for (int i = prevFront + 1; i < top; i++)
            printf("%d ", queue[i]->data);
    }
    else {
        for (int i = top - 1; i > prevFront; i--)
            printf("%d ", queue[i]->data);
    }
}
  
// Driver code
int main()
{
    struct Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(7);
    root->left->right = newNode(6);
    root->right->left = newNode(5);
    root->right->right = newNode(4);
    levelOrder(root, 7);
  
    return 0;
}

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// Java program to print level order traversal
// in spiral form using a single dequeue
class Solution
{
      
static class Node 
{
    int data;
    Node left, right;
};
  
// A utility function to create a new node
static Node newNode(int data)
{
    Node node = new Node();
    node.data = data;
    node.left = node.right = null;
    return (node);
}
  
// function to print the level order traversal
static void levelOrder( Node root, int n)
{
    // We can just take the size as H+N which
    // implies the height of the tree with the 
    // size of the tree
    Node queue[] = new Node[2 * n];
      
    for(int i = 0; i < 2 * n; i++)
        queue[i] = new Node();
      
    int top = -1;
    int front = 1;
    queue[++top] = null;
    queue[++top] = root;
    queue[++top] = null;
  
    // Node t=root;
    int prevFront = 0, count = 1;
    while (true)
    {
  
        Node curr = queue[front];
  
        // A level separator found
        if (curr == null
        {
  
            // If this is the only item in dequeue
            if (front == top)
                break;
  
            // Else print contents of previous level
            // according to count
            else 
            {
                if (count % 2 == 0)
                {
                    for (int i = prevFront + 1; i < front; i++)
                        System.out.printf("%d ", queue[i].data);
                }
                else 
                {
                    for (int i = front - 1; i > prevFront; i--)
                        System.out.printf("%d ", queue[i].data);
                }
  
                prevFront = front;
                count++;
                front++;
  
                // Insert a new level separator
                queue[++top] = null;
  
                continue;
            }
        }
  
        if (curr.left != null)
            queue[++top] = curr.left;
        if (curr.right != null)
            queue[++top] = curr.right;
        front++;
    }
  
    if (count % 2 == 0
    {
        for (int i = prevFront + 1; i < top; i++)
            System.out.printf("%d ", queue[i].data);
    }
    else 
    {
        for (int i = top - 1; i > prevFront; i--)
            System.out.printf("%d ", queue[i].data);
    }
}
  
// Driver code
public static void main(String args[])
{
    Node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(7);
    root.left.right = newNode(6);
    root.right.left = newNode(5);
    root.right.right = newNode(4);
    levelOrder(root, 7);
}
}
  
// This code is contributed by Arnab Kundu

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Output:

1 2 3 4 5 6 7

Time Complexity: O(n)
Auxiliary Space : O(2*n) = O(n)



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