[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 

 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;
        
    }
    
    
};