边工作边刷题：70天一遍leetcode: day 78

Graph Valid Tree

要点：本身题不难，关键是这题涉及几道关联题目，要清楚之间的差别和关联才能解类似题：isTree就比isCycle多了检查连通性，所以这一系列题从结构上分以下三部分

given a graph or edges？有可能给点的编号，然后只给edges，这个最适合用union-find。如果用dfs，那么要先转化成adjacent list表示。这样如果直接给的连通关系的graph，那么dfs就比较直接了
isCycle：directed or undirected：dfs过程中，如果不需要连通，那么要dfs外加一层loop。directed要加一个recStack（一般是个set，因为要判断是不是已经在里面）：visited但是不在stack里，说明是从其他通路曾经访问的，可以早return，但不表示false。而undirected只需要visited，需要检查的只有visited的neighbor不是parent，那么return false。
isTree：directed or undirected：
- dfs，undirected：由于需要连通的特性，undirected比较简单，不需要外层loop，直接一遍dfs后看还有没有没连通的。
- dfs，directed：先要找到root（也就是过一遍所有的结点，找到唯一的无入度的点）。然后从这个结点开始做和undirected一样的dfs，看有没有环，以及最后是否遍历了。注意这里比较tricky，因为如果一个结点有2+个入度，那么一定不是树，而只有这种情况才会出现directed graph无环的特殊情况。
- 而如果用union-find，undirected/directed都类似，就是看最后是不是只有一个root

# 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.

# For example:

# 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.

# Hint:

# Given n = 5 and edges = [[0, 1], [1, 2], [3, 4]], what should your return? Is this case a valid tree?
# According to the definition of tree on Wikipedia: “a tree is an undirected graph in which any two vertices are connected by exactly one path. In other words, any connected graph without simple cycles is a tree.”
# Note: 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.

# Hide Company Tags Google Facebook Zenefits
# Hide Tags Depth-first Search Breadth-first Search Graph Union Find
# Hide Similar Problems (M) Course Schedule (M) Number of Connected Components in an Undirected Graph

class Solution(object):
    def validTree(self, n, edges):
        """
        :type n: int
        :type edges: List[List[int]]
        :rtype: bool
        """
        parents = range(n)
        self.count = n
        def find(x):
            if parents[x]!=x:
                parents[x]=find(parents[x])
            
            return parents[x]
        
        def union(x,y):
            x,y = find(x),find(y)
            if x!=y:
                parents[x]=y
                self.count-=1
                return True
            return False
        
        for e in edges:
            if not union(e[0], e[1]):
                return False
        
        return self.count==1

sol = Solution()
assert sol.validTree(5,  [[0, 1], [0, 2], [0, 3], [1, 4]])==True
assert sol.validTree(5,  [[0, 1], [1, 2], [2, 3], [1, 3], [1, 4]])==False