310. Minimum Height Trees

For an 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 :

Input: n = 4, edges = [[1, 0], [1, 2], [1, 3]]

        0
        |
        1
       / \
      2   3 

Output: [1]
Example 2 :

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

     0  1  2
      \ | /
        3
        |
        4
        |
        5 

Output: [3, 4]
Note:

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.”
The height of a rooted tree is the number of edges on the longest downward path between the root and a leaf.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/minimum-height-trees
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要找的节点是这个多叉树里面的最长路径的中间一个或两个节点，用多源bfs算法，首先找到所有边缘的节点，即入度为零的节点，然后每次遍历都把剩下的边缘节点都删除掉，删除到最后，就是我们的要的节点。

class Solution {
    public List<Integer> findMinHeightTrees(int n, int[][] edges) {
        boolean dp[][] = new boolean[n][n];

        List <Integer> ans = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                dp[i][j] = false;
            }
            ans.add(i);
        }
        if (ans.size() <= 2) {
            return ans;
        }
        // 计算所有节点的入度
        int[] count = new int[n];
        for (int i = 0; i < edges.length; i++) {
            dp[edges[i][0]][edges[i][1]] = true;
            dp[edges[i][1]][edges[i][0]] = true;
            count[edges[i][0]]++;
            count[edges[i][1]]++;
        }
        // 找到入度为1的边缘节点
        List<Integer> inter = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            if (count[i] == 1) {
                inter.add(i);
            }
        }

        while (!inter.isEmpty()) {
            for (int i : inter) {
                for (int j = 0; j < n; j++) {
                    if (dp[i][j]) {
                        count[j]--;
                    }
                }
                count[i]--;
                ans.remove((Integer) i);
            }

            inter.clear();
            for (int j = 0; j < n; j++) {
                if (count[j] == 1) {
                    inter.add(j);
                }
            }
            if (ans.size() <= 2) {
                return ans;
            }
        }
        return ans;
    }
}