坑点太多了

// 除了根结点（root）外，每个结点 “有且只有” 一个父结点，root没有父结点。
// 可以用入度表，每个结点的入度的值是0或1，除此之外就是不合法的树。
// 本题的测试用例包含了多个根节点(多颗树)的情况。（那就是一个森林了forest！）要特判！
#include <iostream>
#include <vector>
#include <queue>
#include <unordered_map>
#include <unordered_set>
#include <sstream>
#include <algorithm>

using namespace std;
int n, parent, child, seqId;

unordered_map<int, int> sequences;
unordered_map<int, int> indegrees;
unordered_map<int, vector<int>> graph; // adjacency list (directed graph)

string levelOrder(int root) {
  queue<int> q{{root}};

  ostringstream out;
  while (not q.empty()) {
    vector<int> tmp;
    size_t sz = q.size();
    while (sz--) {
      const int cur = q.front(); q.pop();
      tmp.emplace_back(cur);
      for (const auto& child : graph[cur])
        q.emplace(child);
    }
    sort(begin(tmp), end(tmp), [](const auto& a, const auto& b) {
      return sequences[a] < sequences[b];
    });
    for (const auto& x : tmp) out << x << ',';
  }
  return out.str();
}

int main(void) {
  scanf("%d", &n);
  while (n--) {
    scanf("%d,%d", &parent, &child);

    auto children = graph[parent]; // deduplcation
    if (find(begin(children), end(children), child) != end(children))
      continue;

    if (++indegrees[child] > 1)
      return puts("Not a tree"), 0;

    graph[parent].emplace_back(child);

    if (!sequences.count(parent))
      sequences[parent] = seqId++;
    if (!sequences.count(child))
      sequences[child] = seqId++;
  }

  int root = -1, root_count = 0; // root_count 为根节点的数量
  for (const auto& p : sequences) {
    root_count += !indegrees[p.first];
    if (!indegrees[p.first]) root = p.first;
  }

  if (root_count != 1) return puts("Not a tree"), 0;

  auto ans = levelOrder(root);
  ans = ans.substr(0, ans.length() - 1);
  printf("%s\n", ans.c_str());
}