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]

题意：

给定一个n个结点n-1条边的无向图（就是树啦），让你找从哪个点出发，到其他结点的最长距离最小？（返回所有答案）

思路：

一开始类似RIP来更新距离，结果TLE。证明O(n^2)的复杂度太大

只好想其他的方法。

答案一定是最长距离的中间结点位置上。

我们要的是中间结点，沿着树的外围每次把叶子结点砍掉，那么，最后剩下的不就是中间结点了么？

# leetcode Minimum Height Trees# http://www.hrwhisper.me/leetcode-minimum-height-trees/class Solution(object):    def findMinHeightTrees(self, n, edges):        """        :type n: int        :type edges: List[List[int]]        :rtype: List[int]        """        if n==1: return [0]        digree = [0 for i in xrange(n)]        g = [[] for i in xrange(n)]        for x,y in edges:            digree[x] += 1            digree[y] += 1            g[x].append(y) #add_edge            g[y].append(x)        leaves = [i for i in xrange(n) if digree[i]==1]        nodes = n        while nodes > 2:            temp = []            for i in leaves:                digree[i] = 0                nodes -= 1                for j in g[i]:                    digree[j] -= 1                    if digree[j] == 1:                        temp.append(j)            leaves = temp        return leaves

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