Given two binary trees, write a function to check if they are the same or not.

Two binary trees are considered the same if they are structurally identical and the nodes have the same value.

Example 1:

Input:     1         1

          / \       / \

         2   3     2   3

        [1,2,3],   [1,2,3]

Output: true

Example 2:

Input:     1         1

          /           \

         2             2

        [1,2],     [1,null,2]

Output: false

Example 3:

Input:     1         1

          / \       / \

         2   1     1   2

        [1,2,1],   [1,1,2]

Output: false
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这个题就是判断两个二叉树是否相等。可以用DFS，也可以用BFS。
emmmmm，按理来说用DFS应该是简单的，至少代码好写，可是，我怎么感觉DFS比BFS还难？？？？

C++代码：

/**

 * Definition for a binary tree node.

 * struct TreeNode {

 *     int val;

 *     TreeNode *left;

 *     TreeNode *right;

 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 * };

 */

class Solution {

public:

    bool isSameTree(TreeNode* p, TreeNode* q) {

        if(p == NULL && q == NULL) return true;

        if(p == NULL || q == NULL) return false;

        if(p->val != q->val) return false;

        return isSameTree(p->left,q->left) && isSameTree(p->right,q->right);

    }

};