leetcode题解Java | 310. Minimum Height Trees

题目：https://leetcode.com/problems/minimum-height-trees/#/description

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:

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.

分析：

解法一：用类型拓扑排序的方法，每次删除度为一的节点，最后剩下的（1个或2个节点），就是root

解法二：找最长路径，最长路径的中点就是root。而最长路径的找法是，先DFS，最长路的末端为起点，再做DFS。注意：递归的时候，不要把大数组放进去，容易栈溢出

Java实现：

public class Solution {    List<Integer>[] lists;public List<Integer> findMinHeightTrees(int n, int[][] edges){lists = new ArrayList[n];for(int i=0; i<n; ++i)lists[i] = new ArrayList<>();for(int i=0; i<edges.length; ++i){int v1 = edges[i][0];int v2 = edges[i][1];lists[v1].add(v2);lists[v2].add(v1);}List<Integer> ans = new ArrayList<>();List<Integer> maxList = DFS(0, n, -1);maxList = DFS(maxList.get(0), n, -1);int len = maxList.size();if(len%2==1)ans.add(maxList.get(len/2));else{ans.add(maxList.get(len/2));ans.add(maxList.get(len/2-1));}return ans;}public List<Integer> DFS(int k, int n, int pre){List<Integer> maxList = new ArrayList<>();for(int i=0; i<lists[k].size(); ++i){int v = lists[k].get(i);if(v!=pre){List<Integer> tmp = DFS(v, n, k);if(tmp.size()>maxList.size())maxList = tmp;}}maxList.add(k);return maxList;}}