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]

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].

判斷一個圖是否是二分圖。這題給出了二分圖的定義：即圖上的頂點可以被分為互相獨立的兩簇，圖上每條的兩個頂點分別在這兩簇中。做法可以是模擬這個定義，對圖上每條邊的頂點進行著色。如果圖上每條邊的頂點都能不沖突的著上兩個色。則可以是二分圖，解法可以是BFS也可以是DFS，下面給出DFS的解法：

class Solution(object):
    def isBipartite(self, graph):
        """
        :type graph: List[List[int]]
        :rtype: bool
        """
        colorset = [-1] * len(graph)
        #防止圖不連通，所以要多次出發。
        for i in xrange(len(graph)):
            if colorset[i] == -1 and not self.validcolor(graph, i, colorset, 0):
                return False
        return True    
        
    def validcolor(self, graph, node, colorset, color):
        colorset[node] = color
        newcolor = 1 - color
        for n in graph[node]:
            if colorset[n] == -1 and not self.validcolor(graph, n, colorset, newcolor):
                return False
            elif colorset[n] != newcolor:
                return False
        return True  

參考解法鏈接：https://leetcode.com/problems/is-graph-bipartite/discuss/115487/Java-Clean-DFS-solution-with-Explanation

785.Is Graph Bipartite?