我认为递归解决方案会更加简洁:
drill (node, clevel, rlevel, result) { if (clevel == rlevel) { result.enqueue (node); else { if (node.left != null) drill (node.left, clevel + 1, rlevel, result); if (node.right != null) drill (node.right, clevel + 1, rlevel, result); }}初始调用如下所示:
drill (root, 0, n, rqueue);



