【LEETCODE】310-Minimum Height Trees

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:

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.

参考： http://bookshadow.com/weblog/2015/11/26/leetcode-minimum-height-trees/ http://www.cnblogs.com/grandyang/p/5000291.html
最先想到的解法是遍历所有的点，以每个点都当做根节点，算出高度，然后找出最小的，但是肯定会Time Limit Exceeded
用类似剥洋葱的方法，就是一层一层的褪去叶节点，最后剩下的一个或两个节点就是我们要求的最小高度树的根节点

class Solution(object):
def findMinHeightTrees(self, n, edges):
"""
:type n: int
:type edges: List[List[int]]
:rtype: List[int]
"""

children=collections.defaultdict(set)     #建立一个图children，是一个二维数组，其中children[i]是一个一维数组，保存了i节点可以到达的所有节点

for s,t in edges:
children[s].add(t)
children[t].add(s)

v=set(children.keys())    #类似剥洋葱的方法，就是一层一层的褪去叶节点，最后剩下的一个或两个节点就是我们要求的最小高度树的根节点

while len(v)>2:

leaves=[x for x in children if len(children[x])==1]        #开始将所有入度为1的节点(叶节点)都存入到一个list中

for x in leaves:
for y in children[x]:
children[y].remove(x)       #通过图来找到和其相连的节点，将该节点的入度减1，如果该节点的入度减到1了，说明此节点也也变成一个叶节点了，加入队列中
del children[x]
v.remove(x)                                 #目标是删除叶节点

return list(v) if n!=1 else [0]