Question
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]
Hint:
 How many MHTs can a graph have at most?
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.
Solution
Result: Accepted Time: 84 ms
Here should be some explanations.
class Solution {
public:
vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
vector<vector<int> > adj_list(n);
vector<int> counters(n, 0),res;
for (const auto &e : edges) {
adj_list[e.first].push_back(e.second);
adj_list[e.second].push_back(e.first);
++counters[e.first];++counters[e.second];
}
queue<int> q;
for (int i = 0; i < n; ++i)
if (counters[i] <= 1)
q.push(i);
while (n > 2) {
const int num_leafs = q.size();
n = num_leafs;
for (int i = 0; i < num_leafs; ++i) {
int node = q.front();q.pop();
for (const auto & neighbor : adj_list[node])
if (counters[neighbor] == 1)
q.push(neighbor);
}
}
while (!q.empty()) res.push_back(q.front()),q.pop();
return res;
}
};
Complexity Analytics
 Time Complexity:
 Space Complexity: