[Swift]LeetCode310. 最小高度树 | 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:,0
        |
        1
       /       2   3 

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

Example 2 :

Input:,0  1  2
       | /
        3
        |
        4
        |
        5 

Output: n = 6edges = [[0,3],[2,[4,[5,4]][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.

对于一个具有树特征的无向图，我们可选择任何一个节点作为根。图因此可以成为树，在所有可能的树中，具有最小高度的树被称为最小高度树。给出这样的一个图，写出一个函数找到所有的最小高度树并返回他们的根节点。

格式

该图包含?n?个节点，标记为?0?到?n - 1。给定数字?n?和一个无向边?edges?列表（每一个边都是一对标签）。

你可以假设没有重复的边会出现在?edges?中。由于所有的边都是无向边，?[0,1]和?[1,0]?是相同的，因此不会同时出现在?edges?里。

示例 1:

输入:,0
        |
        1
       /       2   3 

输出: 
n = 4edges = [[1,3]][1]

示例 2:

输入:,0  1  2
       | /
        3
        |
        4
        |
        5 

输出: n = 6edges = [[0,4]

说明:

• ?根据树的定义，树是一个无向图，其中任何两个顶点只通过一条路径连接。 换句话说，一个任何没有简单环路的连通图都是一棵树。
• 树的高度是指根节点和叶子节点之间最长向下路径上边的数量。

1724 ms

 1 class Solution {
 2     func findMinHeightTrees(_ n: Int,_ edges: [[Int]]) -> [Int]{
 3         var graph = [Int : Set<Int>]()
 4         for edge in edges {
 5             if graph[edge[0]] == nil {
 6                 graph[edge[0]] = [edge[1]]
 7             }else {
 8                 graph[edge[0]]!.insert(edge[1])
 9             }
10             if graph[edge[1]] == nil {
11                 graph[edge[1]] = [edge[0]]
12             }else {
13                 graph[edge[1]]!.insert(edge[0])
14             }
15             
16         }
17         while graph.count > 2 {
18             let list = graph.filter{$1.count == 1}
19             for dict in list {
20                 graph[dict.value.first!]?.remove(dict.key)
21                 graph[dict.key] = nil
22             }
23         }
24         
25         let res = Array(graph.keys)
26         if res.isEmpty {
27             return [0]
28         }
29         return res
30     }
31 }