# Level Order Traversal, Print each level in separate line.

Objective: Given a Binary tree , Print each level of a tree in separate line.

NOTE : This problem is very similar ” Create Linked Lists of all the nodes at each depth “

Input: A binary tree

Output: Each level of binary tree, in one line

Example:

Level Order Traversal, Print each level in one line.

Approach:

Naive Approach:

1. Get the height of the tree.
2. Put a for loop for each level in tree.
3. for each level in step 2, do pre order traversal and print only when height matches to the level.
4. Look at the code for better explanation

Time Complexity : O(N^2) – because each level you are traversing the entire tree.

Better Solution :   Time Complexity – O(N)

• Create a ArrayList of Linked List Nodes.
• For getting all the nodes at each level, before you take out a node from queue, store the size of the queue in a variable, say you call it as levelNodes.
• Now while levelNodes>0, take out the nodes and print it and add their children into the queue.
• After this while loop put a line break.
```while(!q.isEmpty()){
levelNodes = q.size();
while(levelNodes>0){
Node n = (Node)q.remove();
System.out.print(" " + n.data);
levelNodes--;
}
System.out.println("");
}

```
• Since we had taken the queue size before we add new nodes, we will get the count at each level and after printing this count, put a line break, see the example below

Level Order Traversal, Print each level in separate line.

Complete Code:

Output:

```Output by Naive Approach :
5
10 15
20 25 30 35
Output by Better Approach :
5
10 15
20 25 30 35
```

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If you find anything incorrect or you feel that there is any better approach to solve the above problem, please write comment.
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• http://abhis.ws Abhishek

Or you can take two counters, current and next level. Keep printing on the same line till current reaches 0. at that point you have completed a level. make current=next, next=0 and print a line break. this will avoid a second loop.

• Sumit

Did you try implementing it? what values will you store in counters and more importantly how will you know that you have reached to the last node at particular level?? you will need an identifier to insert in queue to know about the level change.

• http://abhis.ws Abhishek

https://gist.github.com/adeydas/d209dca13927fb8a63fd – I checked with only the input in this posts, so feel free to check some more.

• Sumit

Thanks for the link. I have checked the code. Both has the same complexity, what you have said is right. the code will iterate the same number of time.

• Akash G

what is the time complexity of the better solution?

• tutorialhorizon

We are visiting each node only once , so it will be O(N)