单选题

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(   )。

A

1 2 8 9 4 5 3 6 7 10 11

B

1 2 3 4 5 6 7 8 9 10 11

C

1 2 3 8 9 6 4 5 7 10 11

D

1 2 3 8 9 4 5 6 7 10 11

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