[LeetCode] Minimum Height Trees 最小高度树

时间:2022-06-19 23:31:16

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 = 4edges = [[1, 0], [1, 2], [1, 3]]

        0
|
1
/ \
2 3

return [1]

Example 2:

Given n = 6edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]

     0  1  2
\ | /
3
|
4
|
5

return [3, 4]

Hint:

  1. 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 == ) return {};
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 = ; i < n; ++i) {
if (adj[i].size() == ) q.push(i);
}
while (n > ) {
int size = q.size();
n -= size;
for (int i = ; i < size; ++i) {
int t = q.front(); q.pop();
for (auto a : adj[t]) {
adj[a].erase(t);
if (adj[a].size() == ) 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:

Course Schedule II

Course Schedule

Clone Graph

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

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