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 AandBsuch that every edge in the graph has one node inAand 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 nodesiandjexists. Each node is an integer between0and graph.length - 1. There are no self edges or parallel edges:graph[i]does not containi, 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.

leetcode给我们标识了这是一道DFS搜索的题目，那么我们来解读下题目。
给我们一张无向图， 有n个节点，如果这个无向图可以被分为两个子图，使得原来的无向图graph的没条边的两个端点都可以被划分到这两个子图中，如若不然，return false.
依然使用染色法，

• -1 没有颜色
• 1 代表一种颜色
• 0 代表另一种颜色
  因为只需保证相邻端点颜色不同即可， 所以在DFS一个定点的时候，对他的相邻节点使用1 - color来继续DFS
  从第一个顶点开始，如果已经有颜色了，就返回。

class Solution {
    public boolean isBipartite(int[][] graph) {
        int n = graph.length;
        int[] colors = new int[n];
        Arrays.fill(colors, -1);            

        for (int i = 0; i < n; i++) {             
            if (colors[i] == -1 && !validColor(graph, colors, 0, i)) {
                return false;
            }
        }
        return true;
    }

    public boolean validColor(int[][] graph, int[] colors, int color, int node) {
        if (colors[node] != -1) {
            return colors[node] == color;
        }       
        colors[node] = color;       
        for (int next : graph[node]) {
            if (!validColor(graph, colors, 1 - color, next)) {
                return false;
            }
        }
        return true;
    }
}