在普通二叉树(节点值无大小规律)中用 BFS 判断是否存在值为 x 的结点,横线处应填入:
while (!q.empty()) {
TreeNode* cur = q.front(); q.pop();
if (cur->val == x) return cur;
________
}
- A. q.push(cur);
- B. if (cur->right) q.push(cur->right);
- C. if (cur->left) q.push(cur->left); if (cur->right) q.push(cur->right);
- D. q.push(cur->left); q.push(cur->right);
正确答案:C