leetcode 310. Minimum Height Trees

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]

Note:

(1) According to the definition of tree on Wikipedia: “a tree is an undirected graph in which any two vertices are connected by exactlyone 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:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

最开始想把每个点作为根，然后用BFS来按照层数搜索，这样能求出深度。当然是TLE。

class Solution {
public:
    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) 
    {
        vector<vector<int>> map(n, vector<int>(n, 0));
        for (auto it : edges)
        {
            map[it.first][it.second] = 1;
            map[it.second][it.first] = 1;
        }
        vector<int> ret;
        int min_height = INT_MAX;
        
        for (int i = 0; i < n; i++)
        {
            queue<int> que;
            set<int> hash;
            que.push(i);
            hash.insert(i);
            int now_height = 0;
            
            while (!que.empty())
            {
                now_height ++;
                int size = que.size();
                for (int j = 0; j < size; j++)
                {
                    int k = que.front();
                    que.pop();
                    for (int h = 0; h < n; h++)
                    {
                        if (map[k][h] == 1 && hash.find(h) == hash.end())
                        {
                            que.push(h);
                            hash.insert(h);    
                        }
                    } 
                }
                if (now_height > min_height)
                    break;
            }
            
            if (now_height < min_height)
            {
                ret.clear();
                ret.push_back(i);
                min_height = now_height;
            }
            else if (now_height == min_height)
            {
                ret.push_back(i);
            }
        }
        return ret;
    }
};

然后想到用入度的概念来，将入度为1的删除，最后剩下的就是最中间的。

我先用vector<vector<int>>来存相互关系，但是会MLE

所以改成vector<set<int>> 来存。

class Solution {
public:
    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) 
    {
        vector<set<int>> map(n, set<int>());
        vector<int> indegree(n, 0);
        for (auto it : edges)
        {
            map[it.first].insert(it.second);
            map[it.second].insert(it.first);
            indegree[it.first]++;
            indegree[it.second]++;
        }
        vector<int> tmp_ret;
        int left_point = n;
        while (1)
        {
            if ((find_indegree_one(indegree, tmp_ret, left_point)) == 1)
            {
                return tmp_ret;
            }
            left_point -= tmp_ret.size();
            for (int i = 0; i < tmp_ret.size(); i++)
            {    
                for(auto it = map[tmp_ret[i]].begin(); it != map[tmp_ret[i]].end(); it++)
                {
                    indegree[*it]--;
                    map[*it].erase(tmp_ret[i]);
                }
            }
        }
        return tmp_ret;
    }
private:    
    int find_indegree_one(vector<int> &indegree, vector<int> &tmp_ret, int left_point)
    {
        tmp_ret.clear();
        if (left_point == 1) //只剩一个点，肯定是入度为0，其他都为-1
        {
            for (int i = 0; i < indegree.size(); i++)
            {
                if (indegree[i] == 0)
                {
                    tmp_ret.push_back(i);
                    return 1; //返回1 说明找到结果
                }
            }   
        }
        
        for (int i = 0; i < indegree.size(); i++)
        {
            if (indegree[i] == 1)
            {
                tmp_ret.push_back(i);
                indegree[i] = -1;
            }
        }
        if (left_point == 2) //如果还有两个点，肯定是入度为1，其他为0.也返回1结束
            return 1;
        return 0;
    }
};