令n是树的节点数目，下列代码实现了树的广度优先遍历，其时间复杂度是（ ）。

Python
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1from collections import deque
2,[object Object],
Python
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1def bfs(root):,[object Object],
2if not root:,[object Object],
3return,[object Object],
4q = deque(),[object Object],
5q.append(root),[object Object],
6while q:,[object Object],
7node = q.popleft(),[object Object],
8print(node.val, end=" "),[object Object],
9if node.left:,[object Object],
10q.append(node.left),[object Object],
11if node.right:,[object Object],
12q.append(node.right),[object Object],
13```,[object Object],