阅读以下广度优先搜索的代码:

void bfs(TreeNode* root) {   if (root == NULL) {      return;   }   queue<TreeNode*> q;   q.push(root);   while (!q.empty()) {      TreeNode* current = q.front();      q.pop();      cout << current->val << " ";      if (current->left) {         q.push(current->left);      }      if (current->right) {         q.push(current->right);      }   }}

使用以上算法遍历以下这棵树,可能的输出是( )。

    1   / \  2  3 / \  \ 8  9 6   / \ \  4  5  7       / \      10 11
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