310. Minimum Height Trees -- 找出无向图中以哪些节点为根，树的深度最小

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]

Show Hint

Note:

(1) 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.”

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

 

 

vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
    //corner case
    if ( n <=  ) return {};

    //construct a edges search data stucture
    vector<unordered_set<int>> graph(n);
    for (auto e : edges) {
        graph[e.first].insert(e.second);
        graph[e.second].insert(e.first);
    }

    //find all of leaf nodes
    vector<int> current;
    for (int i=; i<graph.size(); i++){
        if (graph[i].size() == )  current.push_back(i);
    }

    // BFS the graph
    while (true) {
        vector<int> next;
        for (int node : current) {
            for (int neighbor : graph[node]) {
                graph[neighbor].erase(node);
                if (graph[neighbor].size() == ) next.push_back(neighbor);
            }
        }
        if (next.empty()) break;
        current = next;
    }
    return current;
}

每次去掉最外面一圈叶子节点。最后剩下的节点集合即为所求。