LeetcodeMinimum Height Trees

Posted 2020-07-10 wuezs

题目链接：https://leetcode.com/problems/minimum-height-trees/

题目：
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]

思路：
多画几个图，直觉最多2种MHT，即最长路径的中心点最多两个。
类似于拓扑排序，从度为1的结点开始一层层剥离，直到最中心两个或一个点。

算法：

    public List<Integer> findMinHeightTrees(int n, int[][] edges) {
        HashMap<Integer, List<Integer>> maps = new HashMap<Integer, List<Integer>>();
        int[] digree = new int[n];// 出入度
        // 记录图的拓扑
        for (int i = 0; i < edges.length; i++) {
            digree[edges[i][0]]++;
            digree[edges[i][1]]++;
            List<Integer> list;

            if (!maps.containsKey(edges[i][0])) {
                list = new ArrayList<Integer>();
            } else {
                list = maps.get(edges[i][0]);
            }
            list.add(edges[i][1]);
            maps.put(edges[i][0], list);

            if (!maps.containsKey(edges[i][1])) {
                list = new ArrayList<Integer>();
            } else {
                list = maps.get(edges[i][1]);
            }
            list.add(edges[i][0]);
            maps.put(edges[i][1], list);
        }

        //获取度为1的点 即叶子节点
        List<Integer> leaves = new ArrayList<Integer>();
        for (int i = 0; i < n; i++) {
            if (digree[i] == 1)
                leaves.add(i);
        }
        int num = n;//图中剩余结点
        List<Integer> tmp;//记录遍历过程中新产生的叶结点
        //沿着叶结点一圈圈剥开图
        while (num > 2) {
            tmp = new ArrayList<Integer>();
            for (int leave : leaves) {
                num--;
                digree[leave] = 0;
                List<Integer> connectedLeave = maps.get(leave);
                for (int c : connectedLeave) {
                    digree[c]--;//删除跟叶结点相邻的边 并将邻结点度数减一
                    if (digree[c] == 1)
                        tmp.add(c);
                }
            }
            leaves = tmp;
        }
        return leaves;
    }