leetcode之Minimum Height Trees

题目：

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]

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.

Credits:

解答：

直接利用BFS的层次搜索即可

class Solution { public:  vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {    // Initialize the undirected graph    vector<unordered_set<int>> adj(n);    for (pair<int, int> p : edges) {      adj[p.first].insert(p.second);      adj[p.second].insert(p.first);    }    // Corner case    vector<int> current;    if (n == 1) {      current.push_back(0);      return current;    }    // Create first leaf layer    for (int i = 0; i < adj.size(); ++i) {      if (adj[i].size() == 1) {        current.push_back(i);      }    }    // BFS the graph    while (true) {      vector<int> next;      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;    }  }};