Leetcode 261: Graph Valid Tree

Posted 2020-10-18 Keep walking

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.

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.

 

  1 public class Solution {
  2     public bool ValidTree(int n, int[,] edges) {
  3         var adjacentList = new IList<int>[n];
  4 
  5         // we need to add the edges twice as the graph is undirected and if we don‘t add twice, we won‘t be able traverse it.
  6         // But when we do BFS/DFS, we need to make sure to traverse the same edge more than once
  7         for (int i = 0; i < edges.GetLength(0); i++)
  8         {
  9             if (adjacentList[edges[i, 0]] == null)
 10             {
 11                 adjacentList[edges[i, 0]] = new List<int>();
 12             }
 13             
 14             adjacentList[edges[i, 0]].Add(edges[i, 1]);
 15             
 16             if (adjacentList[edges[i, 1]] == null)
 17             {
 18                 adjacentList[edges[i, 1]] = new List<int>();
 19             }
 20             
 21             adjacentList[edges[i, 1]].Add(edges[i, 0]);
 22         }
 23 
 24         var visited = new int[n];
 25         var queue = new Queue<int>();
 26         queue.Enqueue(0);
 27         
 28         // there are three states of visited, 0 means not visited, 1 means visiting, 2 means visited
 29         visited[0] = 1;
 30         
 31         while (queue.Count > 0)
 32         {
 33             var cur = queue.Dequeue();
 34             
 35             if (adjacentList[cur] != null)
 36             {
 37                 foreach (var c in adjacentList[cur])
 38                 {
 39                     if (visited[c] == 1) return false;
 40                     if (visited[c] == 0)
 41                     {
 42                         visited[c] = 1;
 43                         queue.Enqueue(c);
 44                     }
 45                 }
 46             }
 47             
 48             visited[cur] = 2;
 49         }
 50         
 51         for (int i = 0; i < n; i++)
 52         {
 53             if (visited[i] == 0) return false;
 54         }
 55         
 56         return true;
 57     }
 58     
 59     private bool HasCycle(IList<int>[] adjacentList, int cur, int prev, bool[] visited)
 60     {
 61         if (visited[cur]) return true;
 62         visited[cur] = true;
 63         
 64         if (adjacentList[cur] != null)
 65         {
 66             foreach (var e in adjacentList[cur])
 67             {
 68                 // since this is undirected graph, we add edges twice so we don‘t want to traverse one edge more than once
 69                 if (e != prev)
 70                 {
 71                     if (visited[e] || HasCycle(adjacentList, e, cur, visited))
 72                     {
 73                         return true;
 74                     }
 75                 }
 76             }
 77         }
 78         
 79         return false;
 80     }
 81 }
 82 
 83 // dfs
 84 public class Solution1 {
 85     public bool ValidTree(int n, int[,] edges) {
 86         var adjacentList = new IList<int>[n];
 87 
 88         for (int i = 0; i < edges.GetLength(0); i++)
 89         {
 90             if (adjacentList[edges[i, 0]] == null)
 91             {
 92                 adjacentList[edges[i, 0]] = new List<int>();
 93             }
 94             
 95             adjacentList[edges[i, 0]].Add(edges[i, 1]);
 96             
 97             if (adjacentList[edges[i, 1]] == null)
 98             {
 99                 adjacentList[edges[i, 1]] = new List<int>();
100             }
101             
102             adjacentList[edges[i, 1]].Add(edges[i, 0]);
103         }
104 
105         var visited = new bool[n];
106         
107         if (HasCycle(adjacentList, 0, -1, visited))
108         {
109             return false;
110         }
111         
112         for (int i = 0; i < n; i++)
113         {
114             if (!visited[i]) return false;
115         }
116         
117         return true;
118     }
119     
120     private bool HasCycle(IList<int>[] adjacentList, int cur, int prev, bool[] visited)
121     {
122         if (visited[cur]) return true;
123         visited[cur] = true;
124         
125         if (adjacentList[cur] != null)
126         {
127             foreach (var e in adjacentList[cur])
128             {
129                 // since this is undirected graph, we add edges twice so we don‘t want to traverse one edge more than once
130                 if (e != prev)
131                 {
132                     if (visited[e] || HasCycle(adjacentList, e, cur, visited))
133                     {
134                         return true;
135                     }
136                 }
137             }
138         }
139         
140         return false;
141     }
142 }