Steps / Pseudo Code:

1. Push all the nodes that are towards the left of tree into a queue; starting from root node all the way down to the the leaf node.
2. Perform the following steps in a loop until queue is empty
   • Pop the node from the queue. Lets call it node P
   • Visit node P
   • Push all the nodes that are towards left of P.right subtree into the queue.
3. That's it.

Points to note:

• Since it's a pre-order traversal, we need to visit the root node first. That justifies the purpose of using a queue. Being a FIFO based data structure, it lets me push the root node first and retrieve it back in the same order (i.e. order of insertion).
• The idea is to push all the nodes that are towards the left in the binary tree to the queue. Whenever a node is visited or popped out from the queue, we need to make sure that right subtree of that node is not left out. That's the point when we push all the nodes, that are towards the left in the right subtree of recently popped out node, into a queue

package info.codeaddict.blog.tree.binary;

import java.util.LinkedList;
import java.util.Queue;

/**
 * @author codeaddict.info
 */
public class PreOrderWithoutRecursion implements Traversal {

    @Override
    public void traverse(BinaryTree.Node root) {
        Queue<BinaryTree.Node> queue = new LinkedList<>();
        pushAllLeft(queue, root);
        while (!queue.isEmpty()) {
            BinaryTree.Node node = queue.poll();
            System.out.println(node.value());
            pushAllLeft(queue, node.right());
        }
    }

    private void pushAllLeft(Queue<BinaryTree.Node> queue, BinaryTree.Node node) {
        while (node != null) {
            queue.add(node);
            node = node.left();
        }
    }
}