Is Graph Bipartite?

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].

class Solution {
    // Input is adjacency list
    // -1 => Node is not colored yet
    // 0 => lets say color blue
    // 1 => lets say color green
    public boolean isBipartite(int[][] graph) {
        int[] color = new int[graph.length];
        Arrays.fill(color, -1);
        for (int i = 0; i < graph.length; i++)
            // Node is uncolored and the result from this connected component is false
            // i.e => this component is not bi-partite
            if ((color[i] == -1) && !bfs(i, graph, color))
                return false;
        return true;
    }

    private boolean bfs(int node, int[][] graph, int[] color) {
        Queue<Integer> q = new LinkedList<>();
        q.add(node);
        color[node] = 0;
        // coloring current node
        while (!q.isEmpty()) {
            int curr = q.poll();
            for (int neighbor : graph[curr]) {
                // If neighbour has the same color
                if (color[neighbor] == color[curr])
                    return false;
                // If neighbor is uncolored
                if (color[neighbor] == -1) {
                    // coloring with opposite color
                    color[neighbor] = 1 - color[curr];
                    q.add(neighbor);
                }
            }
        }
        return true;
    }
}