Printing a BST in level order

queue.addAtEnd(root);
queue.addAtEnd(stopper);

while(queue.hasElems() && currNode = queue.popFront()) {\
if(currNode != stopper) {
   print currNode;
   queue.addAtEnd(left) if left not null;
   queue.addAtEnd(right) if right not null;
} else {
  print new line;
  if queue.hasElems()
      addAtEnd(stopper);
}
}

other solutions from leetcode :

With two queues :

void printLevelOrder(BinaryTree *root) {
  if (!root) return;
  queue<BinaryTree*> currentLevel, nextLevel;
  currentLevel.push(root);
  while (!currentLevel.empty()) {
    BinaryTree *currNode = currentLevel.front();
    currentLevel.pop();
    if (currNode) {
      cout << currNode->data << " ";
      nextLevel.push(currNode->left);
      nextLevel.push(currNode->right);
    }
    if (currentLevel.empty()) {
      cout << endl;
      swap(currentLevel, nextLevel);
    }
  }
}


With one queue but counters :
void printLevelOrder(BinaryTree *root) {
 if (!root) return;
 queue<BinaryTree*> nodesQueue;
 int nodesInCurrentLevel = 1;
 int nodesInNextLevel = 0;
 nodesQueue.push(root);
 while (!nodesQueue.empty()) {
   BinaryTree *currNode = nodesQueue.front();
   nodesQueue.pop();
   nodesInCurrentLevel--;
   if (currNode) {
     cout << currNode->data << " ";
     nodesQueue.push(currNode->left);
     nodesQueue.push(currNode->right);
     nodesInNextLevel += 2;
   }
   if (nodesInCurrentLevel == 0) {
     cout << endl;
     nodesInCurrentLevel = nodesInNextLevel;
     nodesInNextLevel = 0;
   }
 }
}