给定一棵二叉树,采用 BFS 返回其右视图(右视图中每个节点都是该层最右侧的节点)。下面代码每层用 sz 记录本层节点数,横线处应填写( )。
vector<int> rightSideView(TreeNode* root) {
vector<int> result;
if (!root) return result;
queue<TreeNode*> q;
q.push(root);
while (!q.empty()) {
int sz = q.size();
for (int i = 0; i < sz; ++i) {
TreeNode* node = q.front(); q.pop();
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if (node->left) q.push(node->left);
if (node->right) q.push(node->right);
}
}
return result;
}
- A. if (i == 0) result.push_back(node->val);
- B. if (i == sz - 1) result.push_back(node->val);
- C. result.push_back(q.front()->val);
- D. if (node->right) result.push_back(node->right->val);
正确答案：B