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;
    }
  }
};