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 = 4
, edges = [[1, 0], [1, 2], [1, 3]]
code
0 | 1 / \ 2 3
return [1]
orm
Example 2:blog
Given n = 6
, edges = [[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
本文地址(就是个人新blog):http://www.hrwhisper.me/leetcode-minimum-height-trees/
本文由 hrwhisper 原创发布,转载请点名出处