A24036. 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（ ）。from collections import deque class TreeNode: def __init__(self, x): self.val = x self.left = None self.right = None def bfs(root): if not root: return q = deque(…

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（    ）。

from collections import deque

class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

def bfs(root):
    if not root:
        return
    q = deque()
    q.append(root)

    while q:
        node = q.popleft()
        print(node.val, end=" ")

        if node.left:
            q.append(node.left)
        if node.right:
            q.append(node.right)