Binary tree, print only one level (BFS)

Question

I'd like to know how could i do to print only a certain level of a binary tree. I've read many questions here about BFS but found none about printin just one level.

How should i change a common BFS search to print, lets say, only level 2 of this tree:

   6
  / \
 4   8
/ \ / \
1 5 7  9

Leve 2 would be 1, 5, 7, 9. Thank you!

Answer 1

You need to have a level property on your Node. And then when you traverse on the tree, you ask:

if (level == 2) //or whatever level you wish
{
    ...
}

Here is a good example: Find all nodes in a binary tree on a specific level (Interview Query)

If there is no level on the Node and you can't change it, then you can make it as a global variable in the method you make the checks - not preferable but one more solution. I haven't check this answer in code, but I believe it should be very close to the solution.

e.g:

int level = 0;

     public void PrintOneLevelBFS(int levelToPrint)    
     {      
        Queue q = new Queue();
        q.Enqueue(root); //Get the root of the tree to the queue.

        while (q.count > 0)
        {
            level++; //Each iteration goes deeper - maybe the wrong place to add it (but somewhere where the BFS goes left or right - then add one level).
            Node node = q.DeQueue();

            if (level == levelToPrint)
            {
                ConsoleWriteLine(node.Value) //Only write the value when you dequeue it
            }

            if (node.left !=null)
            {
                q.EnQueue(node.left); //enqueue the left child
            }
            if (n.right !=null)
            {
                q.EnQueue(node.right); //enqueue the right child
            }
         }
     }

Answer 2

Ok i got the answer from a professor for a similar problem.

In a binary search tree, find the lowest number on a certain level (GenericTree and GenericQueue are specific classes of the course. Also i traslated the whole exercise so some things may sound weird or not :P

public int calculateMinimum( BinaryTree<Integer> tree, int n ){
    GenericQueue<BinaryTree<Integer>> queue = new GenericQueue<BinaryTree<Integer>>();
    queue.push(tree);
    queue.push(new BinaryTree<Integer>(-1));
    int nActual = 0; //actual level
    while (!queue.isEmpty()){
        tree = queue.pop();
        if (nActual == n){
            int min = tree.getRootData();
            while (!queue.isEmpty()){
                tree = queue.pop();
                if (!tree.getRootData().equals(-1) && (tree.getRootData()<min))
                    min = tre.getRootData();
            }
            return min;
        }
        if (!tree.getLeftChild().getRootData() == null))
            queue.push(tree.getLeftChild());
        if (!tree.getRightChild().getRootData() == null))
            queue.push(tree.getRightChild());
        if ((tree.getRootData().equals(-1) && (!queue.isEmpty())){
            nActual++;
            queue.push(new BinaryTree<Integer>(-1));
        }
    }
    return -1;
}