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 {

public:

    bool dfs(vector<vector<int> >& graph, vector<int>& state, int i, int color){

        for (int j = ; j<graph[i].size(); j++){

            if (state[graph[i][j]] == ){    //没有遍历到时

                state[graph[i][j]] = -color;    //标记该节点颜色同时继续搜索

                return dfs(graph, state, graph[i][j], -color);

            }

            else if (state[graph[i][j]] == color){    //邻居节点中与该节点颜色相同则返回false

                return false;

            }

        }

        return true;

    }

    bool isBipartite(vector<vector<int>>& graph) {

        int node_num = graph.size();

        vector<int> state(node_num,);

        int result = true;

        for(int i=; i<graph.size(); i++){

            if(state[i]== && !dfs(graph, state, i, ))

                result =  false;

        }

        return result;

    }

};