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;
}
}
此题还有递归的解法(未完待续…)
类似题目：
参考资料：
https://discuss.leetcode.com/topic/30572/share-some-thoughts/2
https://discuss.leetcode.com/topic/67543/java-o-n-solution-with-explanation-dfs-twice-beat-95
LeetCode All in One 题目讲解汇总(持续更新中…)
转载请注明来源于 Grandyang 的博客 (grandyang.com)，欢迎对文章中的引用来源进行考证，欢迎指出任何有错误或不够清晰的表达。可以在下面评论区评论，也可以邮件至 grandyang@qq.com
