leetcode 310. Minimum Height Trees-蒲公英云

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]]

return [1]

Example 2:

Given n = 6, edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]

return [3, 4]

class Solution {
    vector<vector<int>> traverse(int m, map<int, vector<int>>&pairedges,int n,int h)
    {
        vector<vector<int>>que;
        vector<int>visited(n);
        vector<int>aa; 
        aa.push_back(m);
        visited[m] = 1;
        que.push_back(aa);
        while (!que.back().empty())
        {
            vector<int>bb;
            for (int i = 0; i < que.back().size(); i++)
            {
                for (int j =0;j< pairedges[que.back()[i]].size();j++)
                {
                    if (visited[pairedges[que.back()[i]][j]] == 0)
                    {
                        visited[pairedges[que.back()[i]][j]] = 1;
                        bb.push_back(pairedges[que.back()[i]][j]);
                    }
                }
            }
            que.push_back(bb);
            if (que.size() >= h + 1)
            {
                if (que.back().empty())
                    que.pop_back();
                return que;
            }
        }
        if (que.back().empty())
            que.pop_back();
        return que;
    }
public:
    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
        vector<int>re;
        if (edges.empty())
            {
            re.push_back(0);
            return re;
        }
        map<int, vector<int>>pairedges;
        for (int i = 0; i < edges.size(); i++)
        {
            pairedges[edges[i].first].push_back(edges[i].second);
            pairedges[edges[i].second].push_back(edges[i].first);
        }
        vector<vector<int>>que = traverse(edges[0].first, pairedges,n,n);
        int m = que.back()[0];
        que.clear();
        que = traverse(m, pairedges,n,n);
        int minh = que.size()/2+1;
        int s = que.size();
        if (s % 2 == 0)
        {
            for (int i = 0; i < que[s / 2 - 1].size(); i++)
            {
                vector<vector<int>>pp = traverse(que[s / 2 - 1][i], pairedges,n,minh);
                if (pp.size() == minh)
                {
                    re.push_back(que[s / 2 - 1][i]);
                    break;
                }
            }
            for (int i = 0; i < que[s / 2].size(); i++)
            {
                vector<vector<int>>pp = traverse(que[s / 2][i], pairedges,n,minh);
                if (pp.size() == minh)
                {
                    re.push_back(que[s / 2][i]);
                    return re;
                }
            }
        }
        else
        {
            for (int i = 0; i < que[s / 2].size(); i++)
            {
                vector<vector<int>>pp = traverse(que[s / 2][i], pairedges,n,minh);
                if (pp.size() == minh)
                {
                    re.push_back(que[s / 2][i]);
                    return re;
                }
            }
        }
        return re;
    }
};