785. Is Graph Bipartite?

Question:

Given an undirected graph, return true if and only if it is bipartite.

Recall that a graph is bipartite if we can split it’s set of nodes into two independent subsets A and B such that every edge in the graph has one node in A and another node in B.

The graph is given in the following form: graph[i] is a list of indexes j for which the edge between nodes i and j exists. Each node is an integer between 0 and graph.length - 1. There are no self edges or parallel edges: graph[i] does not contain i, and it doesn’t contain any element twice.

Example 1:
Input: [[1,3], [0,2], [1,3], [0,2]]
Output: true
Explanation: 
The graph looks like this:
0----1
|    |
|    |
3----2
We can divide the vertices into two groups: {0, 2} and {1, 3}.

Example 2:
Input: [[1,2,3], [0,2], [0,1,3], [0,2]]
Output: false
Explanation: 
The graph looks like this:
0----1
| \  |
|  \ |
3----2
We cannot find a way to divide the set of nodes into two independent subsets.

Note:

graph will have length in range [1, 100].
graph[i] will contain integers in range [0, graph.length - 1].
graph[i] will not contain i or duplicate values.
The graph is undirected: if any element j is in graph[i], then i will be in graph[j].

Solution:

Method 1: dfs + bopartite

  class Solution {
      private int[][] g;
      private int[] visited;
      public boolean isBipartite(int[][] graph) {
          g = graph;
          visited = new int[graph.length];        
          for(int i = 0; i < graph.length; i++){
              if(visited[i] == 0){
                  if(!dfs(i, 1))  return false;
              }
          }
          return true;
      }
      private boolean dfs(int node, int color){
          if(visited[node] == -color) return false;
          else if(visited[node] == color) return true;
          visited[node] = color;
          int[] neighbours = g[node];
          for(Integer neighbour : neighbours){
              if(!dfs(neighbour, -color)) return false;
          }
          return true;
      }
  }

Method 2: bfs

  class Solution {
      public boolean isBipartite(int[][] graph) {
          int[] visited = new int[graph.length];
          Queue<Integer> queue = new LinkedList<>();
          for(int i = 0; i < graph.length; i++){
              if(visited[i] == 0){
                  visited[i] = 1;
                  queue.offer(i);
                  while(!queue.isEmpty()){
                      int cur = queue.poll();
                      for(Integer next : graph[cur]){
                          if(visited[next] == visited[cur]) return false;
                          else if(visited[next] == 0){
                              visited[next] = -visited[cur];
                              queue.offer(next);
                          }
                      }
                  }
              }
          }
          return true;
      }
  }