## 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:

1. 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<int> counters(n, 0),res;
for (const auto &e : edges) {
++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: $O(n)$
• Space Complexity: $O(n)$