310. Minimum Height Trees

For an 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 :

Input: n = 4, edges = [[1, 0], [1, 2], [1, 3]]

        0
        |
        1
       / \
      2   3 

Output: [1]

Example 2 :

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

     0  1  2
      \ | /
        3
        |
        4
        |
        5 

Output: [3, 4]

Note:

• According to the definition of tree on Wikipedia: “a tree is an undirected graph in which any two vertices are connected by exactly one path. In other words, any connected graph without simple cycles is a tree.”

• The height of a rooted tree is the number of edges on the longest downward path between the root and a leaf.

// BFS Topological Sort
vector<int> findMinHeightTrees(int n, vector<vector<int>>& edges) { // time: O(E + V); space: O(E + V)
    if (n == 1) return vector<int>({0});
    vector<int> res;
    vector<unordered_set<int> > graph(n); // adjacency list
    queue<int> q;
    for (auto& e : edges) {
        graph[e[0]].insert(e[1]);
        graph[e[1]].insert(e[0]);
    }
    for (int i = 0; i < n; ++i) {
        if (graph[i].size() == 1) q.push(i);
    }
    while (n > 2) {
        int size = q.size();
        n -= size;
        for (int i = 0; i < size; ++i) {
            int t = q.front(); q.pop();
            for (int neigh : graph[t]) {
                graph[neigh].erase(t);
                if (graph[neigh].size() == 1) q.push(neigh);
            }
        }
    }
    while (!q.empty()) {
        res.emplace_back(q.front()); q.pop();
    }
    return res;
}