Question

Write a routine to list out the nodes of a binary tree in level-order. List the root, then nodes at depth 1, followed by nodes at depth 2, and so on. You must do this in linear time. Prove your time bound (Java)

Answer #1

**/*If you have any query do comment in the comment
section else like the solution*/**

This can be done using a FIFO queue, in which left and right child of a parent node will be pushed in the queue, each node will be pushed and popped from the queue only once, hence the time complexity will be O(n)

```
void levelOrderTraversal(Node root)
{
if(root == null)
return;
Queue<Node> q =new LinkedList<Node>();
q.add(root);
while(q.size() > 0)
{
Node node = q.peek();
q.remove();
System.out.print(node.data + " ");
if(node.lt != null)
q.add(node.lt);
if(node.rt != null)
q.add(node.rt);
}
}
```

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