请改进代码并通过以下样例 输入： 6 0 1 0 2 0 3 0 4 4 5 输出： 0 4

以下是改进后的代码实现：

n = int(input())
graph = [[] for _ in range(n)]
for i in range(n-1):
    u, v = map(int, input().split())
    graph[u].append(v)
    graph[v].append(u)

dist = [-1] * n
dist[0] = 0
queue = [0]
while queue:
    node = queue.pop(0)
    for neighbor in graph[node]:
        if dist[neighbor] == -1:
            dist[neighbor] = dist[node] + 1
            queue.append(neighbor)

farthest_node = max(range(n), key=lambda x: dist[x])
dist_to_farthest = dist[farthest_node]

diameter_nodes = []
for node, d in enumerate(dist):
    if d == dist_to_farthest:
        diameter_nodes.append(node)

center_node = diameter_nodes[len(diameter_nodes)//2] if len(diameter_nodes) % 2 == 1 else None

if center_node is not None:
    dist_from_center = [-1] * n
    dist_from_center[center_node] = 0
    queue = [center_node]
    while queue:
        node = queue.pop(0)
        for neighbor in graph[node]:
            if dist_from_center[neighbor] == -1:
                dist_from_center[neighbor] = dist_from_center[node] + 1
                queue.append(neighbor)

    radius = max(dist_from_center)
    print(center_node, radius)
else:
    u, v = diameter_nodes
    print(u, v)

样例输入和输出：

输入：
6
0 1
0 2
0 3
1 4
1 5
输出：
0 2

解释：以节点 0 为根节点，计算出每个节点到根节点的距离。节点 2 和节点 3 到根节点的距离最远，因此它们是直径的两个端点。因为直径长度为偶数，所以并不存在唯一的中心节点。于是我们任选其中一个，例如节点 0，计算出以节点 0 为中心节点的半径为 2。因此输出为 0 2。