[LeetCode] Graph Valid Tree

Problem Description:

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:

1. 1. Given n = 5 and edges = [[0, 1], [1, 2], [3, 4]], what should your return? Is this case a valid tree?
   2. 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 inedges.

---

As suggested by the hint, just check for cycle and connectedness in the graph. Both of these can be done via DFS.

The code is as follows.

 1 class Solution {
 2 public:
 3     bool validTree(int n, vector<pair<int, int>>& edges) {
 4         vector<vector<int>> neighbors(n);
 5         for (auto e : edges) {
 6             neighbors[e.first].push_back(e.second);
 7             neighbors[e.second].push_back(e.first);
 8         }
 9         vector<bool> visited(n, false);
10         if (hasCycle(neighbors, 0, -1, visited))
11             return false;
12         for (bool v : visited)
13             if (!v) return false; 
14         return true;
15     }
16 private:
17     bool hasCycle(vector<vector<int>>& neighbors, int kid, int parent, vector<bool>& visited) {
18         if (visited[kid]) return true;
19         visited[kid] = true;
20         for (auto neigh : neighbors[kid])
21             if (neigh != parent && hasCycle(neighbors, neigh, kid, visited))
22                 return true;
23         return false;
24     }
25 };

 

转载于:https://www.cnblogs.com/jcliBlogger/p/4738788.html