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 : “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.

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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
     
      
       >& edges) { 4         vector
       
        
         > 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
         
           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
          
           
            >& neighbors, int kid, int parent, vector
            
             & 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 };