Zigzag Binary Tree Traversal
Given a binary tree, return the zigzag level order traversal of the nodes' values of the given tree. Zigzag traversal means starting from left to right, then right to left for the next level and then again left to right and so on in an alternate manner.

For example:

For the binary tree in the image below

The zigzag traversal will be {{2}, {5, 7}, {2, 6, 9}, {4, 11, 5}}.

Input Format:

The first line contains an integer 'T' which denotes the number of test cases or queries to be run. Then the test cases are as follows. The only line of each test case contains elements in the level order form. The line consists of values of nodes separated by a single space. In case a node is null, we take -1 in its place. So -1 would not be a part of the tree nodes. For example, the input for the tree depicted in the below image will be:

For example taking a tree: 1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1

Explanation :

Level 1 : The root node of the tree is 1 Level 2 : Left child of 1 = 2 Right child of 1 = 3 Level 3 : Left child of 2 = 4 Right child of 2 = null (-1) Left child of 3 = 5 Right child of 3 = 6 Level 4 : Left child of 4 = null (-1) Right child of 4 = 7 Left child of 5 = null (-1) Right child of 5 = null (-1) Left child of 6 = null (-1) Right child of 6 = null (-1) Level 5 : Left child of 7 = null (-1) Right child of 7 = null (-1) The first not-null node(of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on. The input ends when all nodes at the last level are null(-1).

Note :

The above format was just to provide clarity on how the input is formed for a given tree. The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as: 1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1

Output Format:

For each test case, print the zigzag level order traversal of the given binary tree separated by single spaces. Print the output of each test case in a separate line.

Note:

You don’t need to print anything, it has already been taken care of.

Constraints:

1

Question

Zigzag Binary Tree Traversal

Given a binary tree, return the zigzag level order traversal of the nodes' values of the given tree. Zigzag traversal means starting from left to right, then right to left for the next level and then again left to right and so on in an alternate manner.

For example:

For the binary tree in the image below

alt text

The zigzag traversal will be {{2}, {5, 7}, {2, 6, 9}, {4, 11, 5}}.

Input Format:

The first line contains an integer 'T' which denotes the number of test cases or queries to be run. Then the test cases are as follows.

The only line of each test case contains elements in the level order form. The line consists of values of nodes separated by a single space. In case a node is null, we take -1 in its place. So -1 would not be a part of the tree nodes.

For example, the input for the tree depicted in the below image will be:

alt text

For example taking a tree:

1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1

Explanation :

Level 1 :
The root node of the tree is 1

Level 2 :
Left child of 1 = 2
Right child of 1 = 3

Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6

Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)

Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)

The first not-null node(of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.

The input ends when all nodes at the last level are null(-1).

Note :

The above format was just to provide clarity on how the input is formed for a given tree. 
The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:

1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1

Output Format:

For each test case, print the zigzag level order traversal of the given binary tree separated by single spaces.

Print the output of each test case in a separate line.

Note:

You don’t need to print anything, it has already been taken care of.

Constraints:

1 <= T <= 100
0 <= N <= 3000
0 <= val <= 10^5

Where ‘T’ is the number of test cases, and ‘N’ is the total number of nodes in the binary tree, and “val” is the value of the binary tree node

Time Limit: 1 sec

Accepted Answer

Iterative Approach

Our very basic intuition is that we are going to traverse the given binary tree using level order traversal where we just need to modify one thing that for every alternate level we need to reverse the order that level.

Steps are as follows:

Create an empty queue.
For each node, first, the node is visited and then it’s child nodes are added in the queue if they exist(First in first out).
For each level that we are getting while iterating till the size of the queue, reverse the level order for every alternate level.
Finally, return the stored zig-zag tree traversal.

Space Complexity: O(n)Explanation:

O(N), where ‘N’ is the number of nodes in the given binary tree.

We are storing all the nodes for returning the answer; thus, the answer will be the size of ‘N’, and also, we are using the level order traversal. So in the worst case, when the given binary tree will be the balanced binary tree, then the total number of nodes at the bottom level of the tree will be ((N/2)+1). So at the bottom level size of the queue will be the order of O(N). So overall space complexity will be O(N).

Time Complexity: O(n)Explanation:

O(N), where ‘N’ is the number of nodes in the given binary tree.

While traversing, we visit each node of the given binary tree which takes O(N) times. So the overall time complexity will be O(N).

Java (SE 1.8)

/*

    Time Complexity - O(N)

    Space Complexity - O(N)



    Where N is the number of nodes in the Binary Tree.

*/



import java.util.Queue;

import java.util.ArrayList;

import java.util.LinkedList;



public class Solution 

{



    //    To reverse the array/list with in given range

    private static void reverse(ArrayList arr, int start, int end) 

    {

        while (start < end) 

        {

            int temp = arr.get(start);

            arr.set(start, arr.get(end));

            arr.set(end, temp);

            start++;

            end--;

        }

    }



    public static ArrayList zigzagTreeTraversal(TreeNode root) 

    {

        //    Store the result.

        ArrayList result = new ArrayList();



        if (root == null) 

        {

            return result;

        }



        //    Declare the queue of tree nodes.

        Queue > level = new LinkedList>();



        //    Push the root in the queue initially.

        level.add(root);



        //    To keep a check whether we need to reverse the order of the current level or not.

        boolean check = false;



        while (level.isEmpty() == false) 

        {

            //    Size of the current level.

            int n = level.size();



            int start = result.size();



            for (int i = 0; i < n; i++) 

            {

                //    Pop the queue and take the front element as current node.

                TreeNode node = level.remove();



                //    Add node value to the current level.

                result.add(node.data);



                //    Push the left child in the queue.

                if (node.left != null) 

                {

                    level.add(node.left);

                }



                //    Push the right child in the queue.

                if (node.right != null) 

                {

                    level.add(node.right);

                }

            }



            //    Check if we need to reverse this level order or not.

            if (check == true) 

            {

                reverse(result, start, result.size() - 1);

            }



            //    Flip the check

            check = !check;

        }



        //    Return zigzag level order traversal.

        return result;

    }

}

C++ (g++ 5.4)

/*

    Time Complexity: O(N)

    Space Complexity: O(N)



    Where 'N' is the number of nodes in a binary tree.

*/



vector zigzagTreeTraversal(TreeNode *root)

{



    //    Store the result.

    vector result;



    if (root == NULL)

    {

        return result;

    }



    //    Declare the queue of tree nodes.

    queue *> level;



    //    Push the root in the queue initially.

    level.push(root);



    //    To keep a check whether we need to reverse the order of the current level or not.

    bool check = false;



    while (level.empty() == false)

    {

        //    Size of the current level.

        int n = level.size();



        int start = result.size();



        for (int i = 0; i < n; i++)

        {

            //    Pop the queue and take the front element as current node.

            TreeNode *node = level.front();

            level.pop();



            //    Add node value to the current level.

            result.push_back(node->data);



            //    Push the left child in the queue.

            if (node->left != NULL)

            {

                level.push(node->left);

            }



            //    Push the right child in the queue.

            if (node->right != NULL)

            {

                level.push(node->right);

            }

        }



        //    Check if we need to reverse this level order or not.

        if (check == true)

        {

            reverse(result.begin() + start, result.end());

        }



        check = !check;

    }



    //    Return zigzag level order traversal.

    return result;

}