310. Minimum Height Trees

Problem

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]

Solution

有点 topological sort 的思想。

再加上BFS 就好了

class Solution {
    void buildGraph(const vector<pair<int, int>>& edges,  vector<list<int>>& graph, vector<int>& inArr){
        for( auto edge : edges){
            int i =  edge.first, j = edge.second;
            graph[i].push_back(j);
            graph[j].push_back(i);
            
            inArr[i]++;
            inArr[j]++;
        }
    }
public:
    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
        if( n == 0) return vector<int>();
        if( edges.empty()) return vector<int>(1,0);
        
        // build graph and in arr
        vector<int> inArr(n, 0);        
        vector<list<int>> graph(n, list<int>());
        buildGraph(edges, graph, inArr);
        
        // bfs traverse        
        vector<int> curLvl, nextLvl;
        for( int i = 0; i < inArr.size() ; i++) {
            if( inArr[i] == 1) {
                curLvl.push_back(i);
            }
        }
        
        vector<int> rst (curLvl.begin(), curLvl.end());
        while(!curLvl.empty()) {
            int i = curLvl.back(); curLvl.pop_back();
            int j = graph[i].front();
            
            inArr[i]--;
            for( int j : graph[i] ) {
                if ( --inArr[j] == 1 ) {
                    nextLvl.push_back(j);
                }
            }
            if(curLvl.empty()) {
                if(nextLvl.empty()) return rst;
                rst = nextLvl;
                swap(curLvl, nextLvl);
            }
        }
        return rst;
    }
    
};