leetcode Minimum Height Trees

Minimum Height Trees

题目详情：

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]

解题方法：

首先构造无向图，n为节点数。之后找到叶结点，即只有一个邻居的节点，将其加入队列q。进行循环，当n > 2 时，循环。loop：size = q.size(); n-= size; 之后循环size次，每次取队列首，并弹出队列首。对于刚取出的队列首的邻居，该邻居删除与队列首相连的边。当邻居为叶结点时，将其加入队列。这时队列中剩余的点就是能构成最小高度的树的点。

代码详情：

class Solution {
public:
    vector<int> findMinHeightTrees(int n, vector<pair<int, int> >& edges) {
        vector<int> result;
        if (n == 0) {
        return result;
} else if (n == 1) {
result.push_back(0);
return result;
}
vector<unordered_set<int> > graph(n);
// make graph
for (int i = 0; i < edges.size(); i++) {
graph[edges[i].first].insert(edges[i].second);
graph[edges[i].second].insert(edges[i].first);
}
// find leaf node, push in queue
queue<int> q;
for (int i = 0; i < n; i++) {
if (graph[i].size() == 1) {
q.push(i);
}
}
while (n > 2) {
int size = q.size();
            n -= size;
for (int i = 0; i < size; i++) {
                int temp = q.front();
                q.pop();
for (auto node:graph[temp]) {
                    graph[node].erase(temp);
                    if (graph[node].size() == 1) {
   q.push(node);
   }  
                }
}
}
while (!q.empty()) {
result.push_back(q.front());
q.pop();
}
        return result;
    }
    
};