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Last Updated: 31 Dec, 2020

Difficulty: Easy

```
In zigzag order, level 1 is printed from left to right fashion, level 2 is printed from right to left. and level 3 is printed from left to right again, and so on…..
```

```
For the given binary tree
```

```
The zigzag traversal is [1, 4, 3, 5, 2, 7, 6]
```

```
The first line contains an integer 'T' which denotes the number of test cases or queries to be run. Then the 'T' test cases follow.
The first line of each test case contains elements of the tree 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.
For example, the input for the tree depicted in the below image
```

```
1
3 8
5 2 7 -1
-1 -1 -1 -1 -1 -1
```

```
Level 1 :
The root node of the tree is 1
Level 2 :
Left child of 1 = 3
Right child of 1 = 8
Level 3 :
Left child of 3 = 5
Right child of 3 = 2
Left child of 8 =7
Right child of 8 = null (-1)
Level 4 :
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 2 = null (-1)
Right child of 2 = null (-1)
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).
```

```
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 3 8 5 2 7 -1 -1 -1 -1 -1 -1 -1
```

```
For each test case, print a single line containing all the nodes value in zigzag order traversal separated by a single space in a separate line.
```

```
You do not need to print anything, it has already been taken care of. Just implement the given function.
```

```
1 <= 'T' <= 100
0 <= 'N' <= 10^3
0 <= 'NODES' <= 10^9
Where NODES represent any node value
Time Limit: 1sec
```

We can use level order traversal (recursive) to explore all levels of the tree. Also, at each level nodes should be printed in alternating order.

For example - First level of tree should be printed in left to right manner, Second level of tree should be printed in right to left manner, Third again in left to right order and so on.

So, we will use a Direction variable whose value will toggle at each level and we will print levels in alternating order by looking at the direction of current level.

We will first calculate the maximum level of the binary tree, then we will run a loop from 1 to maximum level and store nodes of each level in our answer based on direction.

If direction is 0 then traverse the whole binary tree from left to right and store nodes of current level in left to right fashion.

If direction is 1 then traverse the whole binary tree from right to left and store nodes of current level in right to left fashion.

The idea here is to use two stacks, one to store odd levels and the other to store even level nodes. We will pop all nodes from the odd level stack and add children’s of a popped node from left to right to even level stack. Then we will pop all nodes from the even level stack and add their children from right to left in the odd level stack. As stack works last in fast out order so we will put children in reverse order so that we can get original order.

**Steps :**

- Declare an empty array
**answer**to store zigzag traversal. - Declare two empty stacks for odd level and even level.
- Initially push root to an odd level stack.
- Run a loop until both stacks are not empty and do:
- Run a loop until the odd level stack is not empty and do:
- Pop the top node from it.
- Add data from popped node to the
**answer**. - Add a left child and right child of popped node to even level stack.

- Run a loop until the even level stack is not empty and do:
- Pop the top node from it.
- Add data from popped node to the
**answer**. - Add the right child and left child of the popped node to an odd level stack.

- Run a loop until the odd level stack is not empty and do:
- At last return
**answer.**

The idea here is to use a breadth-first search i.e. **BFS** along with deque.

We will run a **BFS** and if the level is odd then we will pop elements from front and children of a popped node from left to right in back of the deque.

If the level is even then we will pop all nodes from the back and then add children's popped node from right to left in front of deque.

**Steps :**

- Declare an empty array
**answer**to store zigzag traversal. - Declare an empty deque and push back root to it.
- Declare a variable
**level**to get current level and initialize it to 1. - Run a loop until deque is not empty and do:
- Size of deque denotes the number of nodes in the current level.
- Run a loop until total nodes in the current level are greater than zero, and do:
- If the level is odd then pop the front node from deque and add its data to
**answer**. - Push the left and right child of the popped node to deque.
- If the level is even then pop back node from deque and add its data to
**answer**. - Push the right and left child of the popped node to the front of the deque.

- If the level is odd then pop the front node from deque and add its data to
- Increment level by 1.

- Return the
**answer**.

The idea here is to use a breadth-first search i.e. **BFS** along with the queue.

We use an array to store nodes in current level if the level is even then we will add elements of the array in reverse order otherwise we will add elements of the array in the same order.

**Steps :**

- Declare an empty array
**answer**to store zigzag traversal. - Declare an empty queue and push root to it and declare a boolean variable
**reverseOrder**which will tell us whether we should include nodes in current level in the original order or reverse order. - Initialize
**reverseOrder**as false. - Run a loop until the queue is not empty and do:
- Size of queue denoted by variable
**size**i.e the number of nodes in the current level. - Create an array to store elements of the current level, say
**currentLevelNodes**. - Run a loop until nodes in the current level are greater than zero and add all its nodes data to
**currentLevelNodes** - Push left and right child of current level nodes to queue for next-level iteration
- If
**reverseOrder**is false then add elements of**currentLevelNodes**to answer, else add elements of**currentLevelnodes**to**answer**in reverse order. - Change
**reverseOrder**to false if it is true and vice-versa for the next level.

- Size of queue denoted by variable

5. At last, return the **answer**.

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