c++ - Counting Nodes with single child in BST -

i have question : implement function int countsingle(), uses breadth-first traversal (bft) in order count how many nodes in tree have single child.

so code below thought of solve , there way of doing or more efficient missed ?

template <class t>         int bst<t>::countsingle()         {             queuedll<bstnode<t>*> queue;             bstnode<t>* node = root;              // if tree empty, there no nodes traverse.             if(node == null) return -1;              // initially, put root in queue             queue.enqueue(node);             int counter = 0 ;              while(queue.isempty() == false)             {                 // take node out of queue, process , insert                 // children queue.                 node = queue.dequeue();                  // if node has 1 1 child wether left or right , increase counter                 if((node->left == null && node->right != null) || (node->right== null && node->left!=null))                     counter++;                  if(node->left != null)                     queue.enqueue(node->left);                 if(node->right != null)                     queue.enqueue(node->right);             }             return counter;         } 

this seems pretty ok. can't improve complexity of did, since number of operations linear.

a few implementation points:

• counting non-modifying operation, might want consider consting stuff.

• you've implemented on own similar compiler in recursion. if that's intent - great (and there performance points in favor). fyi, using recursion few locs.