In a Binary Tree, Create Linked Lists of all the nodes at each depth.

Objective: Given a Binary tree create Linked Lists of all the nodes at each depth , say if the tree has height k then create k linked lists.

NOTE : This problem is very similar “Print binary tree, each level in one line

Input: A binary tree

Output: K linked lists if the height of tree is k. Each linked list will have all the nodes of each level.


Linked Lists of all the nodes at each depth Example



  • 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. add these to a linked list
  • After this while loop put a line break and create a new linked list
ArrayList al = new ArrayList();
			levelNodes = q.size();
			ListNode head = null;
			ListNode curr = null;
				Node n = (Node)q.remove();
				ListNode ln = new ListNode(;
					head = ln;
					curr = ln;
				}else{ = ln;
					curr =;
				if(n.left!=null) q.add(n.left);
				if(n.right!=null) q.add(n.right);

  • 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
  • Time Complexity : O(N)

Linked Lists of all the nodes at each depth - Implement

Complete Code:



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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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1 Response

  1. September 23, 2014

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