用bfs模拟判断遍历顺序与所给序列是否一致，样例还行，一提交，似乎一个不过。```cpp #include <bits/stdc++.h>

using namespace std;

bool BFS(const vector<vector>& tree, const vector& s) { queue q; unordered_set visited; q.push(1); visited.insert(1); int idx = 0;

while (!q.empty()) {
    int v = q.front();
    q.pop();
    if (s[idx] != v) {
        return false;
    }
    idx++;
    for (int u : tree[v]) {
        if (visited.find(u) == visited.end()) {
            q.push(u);
            visited.insert(u);
        }
    }
}
return true;

} int main() { int n; cin >> n;

vector<vector<int>> tree(n + 1);
for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    tree[u].push_back(v);
    tree[v].push_back(u);
}

vector<int> s(n);
for (int i = 0; i < n; ++i) {
    cin >> s[i];
}

if (BFS(tree, s)) {
    cout << "Yes";
} else {
    cout << "No";
}

// system("pause"); return 0; }