leetcode 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.node

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).python

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.app

Example 1:spa

Given n = 4edges = [[1, 0], [1, 2], [1, 3]]code

        0
        |
        1
       / \
      2   3

return [1]orm

Example 2:blog

Given n = 6edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]leetcode

     0  1  2
      \ | /
        3
        |
        4
        |
        5

return [3, 4]get

题意:it

给定一个n个结点n-1条边的无向图(就是树啦),让你找从哪一个点出发,到其余结点的最长距离最小?(返回全部答案)

思路:

一开始相似RIP来更新距离,结果TLE。证实O(n^2)的复杂度太大

只好想其余的方法。

答案必定是最长距离的中间结点位置上。

咱们要的是中间结点,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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