【Leetcode】Minimum Height Trees

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
;// 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;
}