LeetCode Construct Quad Tree(dfs)

问题：给出一个n*n由0和1组成的二维数组，用四叉树来表示网格。要求返回四叉树的根结点。

（1）如果当前网格值相同，则是叶子结点，将网格值赋值给结点

（2）如果当前网格值不同，则不是叶子结点，结点值可以为True或者False

（3）递归处理子网格作为其子结点

思路：根据定义做就可以了。

具体代码参考：

https://github.com/wuli2496/OJ/tree/master/LeetCode/Construct%20Quad%20Tree

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