Leetcode minimum height trees

題目

For a undirected graph with tree characteristics, we can choose any node as the root. The result graph is then a rooted tree. Among all possible rooted trees, those with minimum height are called minimum height trees (MHTs). Given such a graph, write a function to find all the MHTs and return a list of their root labels.

Format
The graph contains n nodes which are labeled from 0 to n - 1. You will be given the number n and a list of undirected edges (each edge is a pair of labels).

You can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0, 1] is the same as [1, 0] and thus will not appear together in edges.

Example 1:

Given n = 4, edges = [[1, 0], [1, 2], [1, 3]]

    0
    |
    1
   / \
  2   3

return [1]

Example 2:

Given n = 6, edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]

 0  1  2
  \ | /
    3
    |
    4
    |
    5

return [3, 4]

題目分析

挺有趣的一道題目，給出一個undirected的無環圖，找到它最小樹高度的結點集。一開始往DP方向去想，樹的動態規劃沒怎麼學過，只好放棄。另一種比較直接明瞭的辦法就是BFS。

用BFS去尋找葉節點，慢慢將葉節點都去除，直到剩下的結點數連接數小於等於1，則將剩下的結點數返回。

源代碼

class Solution {
public:
    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
        vector<unordered_set<int> > adj(n);

        //建立鄰接矩陣
        for (int i = 0; i < edges.size(); i++) {
            adj[edges[i].first].insert(edges[i].second);
            adj[edges[i].second].insert(edges[i].first);
        }


        vector<int> current;
        //
        if (n == 1)
            current.push_back(0);
        for (int i = 0; i < n; i++) {
            if (adj[i].size() == 1) {
                current.push_back(i);

            }
        }

        while (1) {
            vector<int> next;
            //找出葉節點
            for (int node : current) {
                for (int neighbor : adj[node]) {
                  adj[neighbor].erase(node);
                  if (adj[neighbor].size() == 1) next.push_back(neighbor);
                }
              }
            if (next.empty())   return current;
            current = next;
        }
    }
};