标签：
Given n nodes labeled from 0 to n - 1 and a list of undirected edges (each edge is a pair of nodes), write a function to check whether these edges make up a valid tree.
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.
Given n = 5 and edges = [[0, 1], [0, 2], [0, 3], [1, 4]], return true.
Given n = 5 and edges = [[0, 1], [1, 2], [2, 3], [1, 3], [1, 4]], return false.
首先看下树的定义：详见维基百科,只要没有回路的连通图就是树.
如果无向简单图G有有限个顶点（设为n个顶点），那么G 是一棵树还等价于：G是连通的，有n − 1条边，并且G没有简单回路.
后面这条定义更加关键, 也就是说我们先判断是不是有n-1边，不是则肯定没法构成树．
之后我们判断是否有简单回路．有回路，则肯定所有节点之间不是全部联通的．可以中途退出．
如何判断是否有回路呢，可以使用并查集，在并查集中，所有联通的块都有同一个祖先节点．如果新加入的这条边，两个节点的祖先节点一样，说明已经联通，会构成环，则说明不合格．
并查集大小是n,合并和查找操作O(n)级别的，所以最终代码时间复杂度为O(nlog*n)级别的，也就是Ｏ(n)级别．另外这题顶点已经预先给定，所以用数组是一个比hashmap更节省空间的选择．
代码如下：
class Solution: # @param {int} n an integer # @param {int[][]} edges a list of undirected edges # @return {boolean} true if it‘s a valid tree, or false def validTree(self, n, edges): if not n: return True if n - 1 != len(edges): return False UF = UnionFind(n) for edge in edges: if UF.find(edge[0]) == UF.find(edge[1]): return False UF.union(edge[0], edge[1]) return True class UnionFind(object): def __init__(self, n): self.id = range(n) self.sz = [1] * n def union(self, x, y): # union is carried out between fathers i = self.find(x) j = self.find(y) if self.sz[i] > self.sz[j]: i, j = j, i self.id[j] = i self.sz[i] += self.sz[j] def find(self, i): while i != self.id[i]: self.id[i] = self.id[self.id[i]] i = self.id[i] return i
标签：
原文地址：http://www.cnblogs.com/sherylwang/p/5639001.html
