**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.

**Example:**

**Approach:**

**Recursion:**

- 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();
while(!q.isEmpty()){
levelNodes = q.size();
ListNode head = null;
ListNode curr = null;
while(levelNodes>0){
Node n = (Node)q.remove();
ListNode ln = new ListNode(n.data);
if(head==null){
head = ln;
curr = ln;
}else{
curr.next = ln;
curr = curr.next;
}
if(n.left!=null) q.add(n.left);
if(n.right!=null) q.add(n.right);
levelNodes--;
}
al.add(head);
}

*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)**

**Complete Code:**

**Output**:

->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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