Symmetric Tree

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

    1
   / \
  2   2
 / \ / \
3  4 4  3

But the following is not:

    1
   / \
  2   2
   \   \
   3    3

Note:
Bonus points if you could solve it both recursively and iteratively.

判断树是不是对称的，这边分别利用一个队列来记录下一层节点的情况，然后进行判断。特别注意空树和只有根节点的树这两种特殊情况！！！

还有就是对给的第二个例子的那种情况，左右进入队列都要进行判断！当然，也可以用栈来实现。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
	bool isSymmetric(TreeNode *root)
	{
		if (!root)
			return true;
		queue<TreeNode*>qleft;
		queue<TreeNode*>qright;
		if(root->left)
		   qleft.push(root->left);
		if(root->right)
		   qright.push(root->right);
		if(qleft.size()!=qright.size())
		    return false;
		while (!qleft.empty() && !qright.empty())
		{
			if (qleft.front()->val != qright.front()->val)
				return false;
			if (qleft.front()->left)
				qleft.push(qleft.front()->left);
			if (qright.front()->right)
				qright.push(qright.front()->right);
			if (qleft.size() != qright.size())
				return false;
				
			if (qleft.front()->right)
				qleft.push(qleft.front()->right);
			if (qright.front()->left)
				qright.push(qright.front()->left);
			qleft.pop();
			qright.pop();
			if (qleft.size() != qright.size())
				return false;
		}
		return true;
	}
};