[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]

Hint:

1. How many MHTs can a graph have at most?

Note:

(1) 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.”

(2) The height of a rooted tree is the number of edges on the longest downward path between the root and a leaf.

解法一：

很像topological sort的算法。

class Solution {public:    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {        if(n==1) return {0};        vector<int> in(n,0);        vector<int> res;        vector<vector<int>> graph(n,vector<int>());                for(auto a:edges){            graph[a.first].push_back(a.second);            ++in[a.first];            graph[a.second].push_back(a.first);            ++in[a.second];        }                queue<int> q;        for(int i=0; i<n; i++){            if(in[i]==1) q.push(i);        }                while(n>2){            int sz = q.size();            for(int i=0; i<sz; i++){                int cur = q.front(); q.pop();                --n;                for(auto a:graph[cur]){                    --in[a];                    if(in[a]==1) q.push(a);                }            }        }                while(!q.empty()){            res.push_back(q.front());            q.pop();        }        return res;            }        };