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。还需要一个一维数组d用来保存各个节点的入度信息，其中d[i]表示i节点的入度数。我们的目标是删除所有的叶节点，叶节点的入度为1，所以我们开始将所有入度为1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，将该节点的入度减1，如果该节点的入度减到1了，说明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：
// Non-recursion
class Solution {
public:
vector<int> findMinHeightTrees(int n, vector<pair<int, int> >& edges) {
if (n == 1) return {0};
vector<int> res, d(n, 0);
vector<vector<int> > g(n, vector<int>());
queue<int> q;
for (auto a : edges) {
g[a.first].push_back(a.second);
++d[a.first];
g[a.second].push_back(a.first);
++d[a.second];
}
for (int i = 0; i < n; ++i) {
if (d[i] == 1) q.push(i);
}
while (n > 2) {
int sz = q.size();
for (int i = 0; i < sz; ++i) {
int t = q.front(); q.pop();
--n;
for (int i : g[t]) {
--d[i];
if (d[i] == 1) q.push(i);
}
}
}
while (!q.empty()) {
res.push_back(q.front()); q.pop();
}
return res;
}
};
此题还有递归的解法(未完待续...)
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参考资料：
https://leetcode.com/discuss/71697/share-my-solution-c-1000ms
https://leetcode.com/discuss/71676/share-my-accepted-solution-java-o-n-time-o-n-space
LeetCode All in One 题目讲解汇总(持续更新中...)