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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]
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.
vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) { //corner case if ( n <= 1 ) return {0}; //construct a edges search data stucture vector<unordered_set<int>> graph(n); for (auto e : edges) { graph[e.first].insert(e.second); graph[e.second].insert(e.first); } //find all of leaf nodes vector<int> current; for (int i=0; i<graph.size(); i++){ if (graph[i].size() == 1) current.push_back(i); } // BFS the graph while (true) { vector<int> next; for (int node : current) { for (int neighbor : graph[node]) { graph[neighbor].erase(node); if (graph[neighbor].size() == 1) next.push_back(neighbor); } } if (next.empty()) break; current = next; } return current; }
每次去掉最外面一圈叶子节点。最后剩下的节点集合即为所求。
310. Minimum Height Trees -- 找出无向图中以哪些节点为根，树的深度最小
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原文地址：http://www.cnblogs.com/argenbarbie/p/5418307.html
