Graph Valid Tree

Givennnodes labeled from0ton-1and 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.

Example 1:

Input:
n = 5
, and 
edges = [[0,1], [0,2], [0,3], [1,4]]
Output:
 true

Example 2:

Input:
n = 5, 
and 
edges = [[0,1], [1,2], [2,3], [1,3], [1,4]]
Output:
 false

Note: you can assume that no duplicate edges will appear inedges. Since all edges are undirected,[0,1]is the same as[1,0]and thus will not appear together inedges.

分析

DFS

建图 graph = {i:[] for i in range(n)}，无方向所以双向

从node 0起for loop，no cycle(dfs) and len(seen) ==n

seen这里做visited，不需要弹出因为树不回头

注意all用法。要排除grandparent!=neighbor是因为双向图，排除2个点互指的情况。

class Solution:
    def validTree(self, n: int, edges: List[List[int]]) -> bool:        
        seen = set()
        graph = {i:[] for i in range(n)}
        for e in edges:#undirected
            graph[e[0]].append(e[1])
            graph[e[1]].append(e[0])
        return self.dfs(graph,seen,0,-1) and len(seen) == n


    def dfs(self, graph, seen,node,parent):
        if node in seen:
            return False
        seen.add(node)
        return all([self.dfs(graph,seen,n,node) for n in graph[node] if n!=parent]) #child!=grandparent

bfs

start from any node and delete every node we meet during the BFS. If there is any node left, then there is an isolated component.

注意dict pop用法，no isolated component =no cycle.

class Solution:
    def validTree(self, n: int, edges: List[List[int]]) -> bool:        
        if len(edges) != n-1: return False #别忘了这个!!!!!!!!
        g,q = {i:[] for i in range(n)},[0]
        for a,b in edges:#undirected
            g[a].append(b)
            g[b].append(a)
        for i in q:
            q+=g.pop(i,[])
        return not g

Union find

union two nodes that are already in the same group, then there is a cycle.

class Solution:
    def validTree(self, n: int, edges: List[List[int]]) -> bool:        
        if len(edges) != n-1: return False #别忘了这个
        f = list(range(n))
        def find(x):
            if f[x]!=x:
                f[x] = find(f[x])
            return f[x]   
        for a,b in edges:
            fa,fb = find(a),find(b)
            if fa==fb:
                return False
            f[fa] = fb
        return True