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:
- 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.
Credits:
Special thanks to @peisi for adding this problem and creating all test cases.
Update (2015-11-25):
The function signature had been updated to return List<Integer>
instead of integer[]
. Please click the reload button above the code editor to reload the newest default code definition.
这道题虽然是树的题目,但是跟其最接近的题目是Course Schedule 课程清单和Course Schedule II 课程清单之二。由于LeetCode中的树的题目主要都是针对于二叉树的,而这道题虽说是树但其实本质是想考察图的知识,这道题刚开始在拿到的时候,我最先想到的解法是遍历的点,以每个点都当做根节点,算出高度,然后找出最小的,但是一时半会又写不出程序来,于是上网看看大家的解法,发现大家推崇的方法是一个类似剥洋葱的方法,就是一层一层的褪去叶节点,最后剩下的一个或两个节点就是我们要求的最小高度树的根节点,这种思路非常的巧妙,而且实现起来也不难,跟之前那到课程清单的题一样,我们需要建立一个图g,是一个二维数组,其中g[i]是一个一维数组,保存了i节点可以到达的所有节点。我们开始将所有只有一个连接边的节点(叶节点)都存入到一个队列queue中,然后我们遍历每一个叶节点,通过图来找到和其相连的节点,并且在其相连节点的集合中将该叶节点删去,如果删完后此节点也也变成一个叶节点了,加入队列中,再下一轮删除。那么我们删到什么时候呢,当节点数小于等于2时候停止,此时剩下的一个或两个节点就是我们要求的最小高度树的根节点啦,参见代码如下:
C++ 解法一:
class Solution { public: vector<int> findMinHeightTrees(int n, vector<pair<int, int> >& edges) { if (n == 1) return {0}; vector<int> res; vector<unordered_set<int>> adj(n); queue<int> q; for (auto edge : edges) { adj[edge.first].insert(edge.second); adj[edge.second].insert(edge.first); } for (int i = 0; i < n; ++i) { if (adj[i].size() == 1) q.push(i); } while (n > 2) { int size = q.size(); n -= size; for (int i = 0; i < size; ++i) { int t = q.front(); q.pop(); for (auto a : adj[t]) { adj[a].erase(t); if (adj[a].size() == 1) q.push(a); } } } while (!q.empty()) { res.push_back(q.front()); q.pop(); } return res; } };
Java 解法一:
public class Solution { public List<Integer> findMinHeightTrees(int n, int[][] edges) { if (n == 1) return Collections.singletonList(0); List<Integer> leaves = new ArrayList<>(); List<Set<Integer>> adj = new ArrayList<>(n); for (int i = 0; i < n; ++i) adj.add(new HashSet<>()); for (int[] edge : edges) { adj.get(edge[0]).add(edge[1]); adj.get(edge[1]).add(edge[0]); } for (int i = 0; i < n; ++i) { if (adj.get(i).size() == 1) leaves.add(i); } while (n > 2) { n -= leaves.size(); List<Integer> newLeaves = new ArrayList<>(); for (int i : leaves) { int t = adj.get(i).iterator().next(); adj.get(t).remove(i); if (adj.get(t).size() == 1) newLeaves.add(t); } leaves = newLeaves; } return leaves; } }
本文转自博客园Grandyang的博客,原文链接:最小高度树[LeetCode] Minimum Height Trees ,如需转载请自行联系原博主。