[LeetCode] Minimum Height Trees 最小高度树

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]]

return [1]

Example 2:

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

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 == ) 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;

    }

}