请实现一个函数，用来判断一颗二叉树是不是对称的。注意，如果一个二叉树同此二叉树的镜像是同样的，定义其为对称的。

#include "stdafx.h"
#include<iostream>
#include<vector>
#include<string>
#include<queue>
#include<stack>

using namespace std;

struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
TreeNode(int x) :
val(x), left(NULL), right(NULL) {
}
};
class Solution {
public:
//层次遍历二叉树
//都从上往下遍历；一个从左往右遍历，一个从右往左遍历；
//如果得到的层次遍历结果一样，说明这棵树是对称二叉树
//上面这种方法不可行，因为比如1 2 # # #和 1 # 2 # #就不是对称二叉树

//方法2：构建镜像二叉树
TreeNode *newTree=NULL;
bool isSymmetrical(TreeNode* pRoot)
{
if (pRoot == NULL) return false;
CreateImageTree(pRoot, newTree);
bool result = judgeTwoTree(pRoot, newTree);
return result;
}
void CreateImageTree(TreeNode* pRoot,TreeNode* &T)
{
if (pRoot == NULL) return;
else
{
int num = pRoot->val;
T = new TreeNode(num);
CreateImageTree(pRoot->left,T->right);
CreateImageTree(pRoot->right,T->left);
}
}
bool judgeTwoTree(TreeNode* T1,TreeNode* T2)
{
if (T1 == NULL && T2 == NULL) return true;
if (T1 != NULL && T2 == NULL) return false;
if (T1 == NULL && T2 != NULL) return false;
if (T1->val != T2->val) return false;
return judgeTwoTree(T1->right, T2->right) && judgeTwoTree(T1->left,T2->left);
}

void preCreate(TreeNode* &T)
{
int num;
cin >> num;
if (num == 0) T = NULL;
else
{
T = new TreeNode(num);
preCreate(T->left);
preCreate(T->right);
}
}

void preOrder(TreeNode* T)
{
if (T == NULL) return;
else
{
cout << T->val << "  ";
preOrder(T->left);
preOrder(T->right);
}
}
};
int main()
{

Solution so;
TreeNode *T1;
TreeNode *T2;
vector<int> vec;

cout << "创建T1:" << endl;
so.preCreate(T1);
cout << "创建T1成功！" << endl;

cout << "T1的前序遍历:" << endl;
so.preOrder(T1);
cout << endl;

//so.CreateImageTree(T1,T2);
//cout << "T2的前序遍历是（T2是T1的镜像）："<<endl;
//so.preOrder(T2);
//cout << endl;

cout << "T1是否是对称的：" ;
bool re = so.isSymmetrical(T1);
cout << re << endl;

cout << endl;
return 0;
}