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.
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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