Be the first user to complete this post

  • 0
 Add to List
Medium

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.
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.
  • Do the level order traversal using queue(Breadth First Search). Click here to know about how to level order traversal.
  • 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);
		if(n.left!=null) q.add(n.left);
		if(n.right!=null) q.add(n.right);
		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.
Level Order Traversal, Print each level in separate line.

Code:


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

Also Read: