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 3return [1]
Example 2:
Given n = 6, edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]
0 1 2 \ | / 3 | 4 | 5return [3, 4]
Show Hint 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.
class Solution { public: vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) { if(n == 1) { vector<int> a(1, 0); return a; } int degree[n]; vector<int> G[n]; memset(degree, 0, sizeof(degree)); vector<int> r; int m = edges.size(); for(int i=0; i<m; i++) { int u = edges[i].first; int v = edges[i].second; degree[u]++; degree[v]++; G[u].push_back(v); G[v].push_back(u); } for(int i=0; i<n; i++) if(degree[i] == 1) r.push_back(i); int cnt = n; while(cnt > 2) { vector<int> tmp; for(int i=0; i<r.size(); i++) { int u = r[i]; cnt -= 1; degree[u] = 0; for(int j=0; j<G[u].size(); j++) { int v = G[u][j]; degree[v]--; if(degree[v] == 1) tmp.push_back(v); } } r = tmp; } return r; } };