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

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

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

But the following [1,2,2,null,3,null,3] is not:

    1
/ \
2 2
\ \
3 3

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

/*
递归实现: 要考虑到左右两个子树去递归,所以重写函数 */
/**
* 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 helper(TreeNode* left, TreeNode* right){
if (left == NULL && right == NULL){
return true;
}else if (left == NULL && right != NULL){
return false;
}else if (left != NULL && right == NULL){
return false;
}else if (left -> val == right -> val){
return helper(left -> left, right -> right) && helper(left -> right, right -> left);
}else{
return false;
}
}
bool isSymmetric(TreeNode* root) {
if (root == NULL){
return true;
}
return helper(root -> left, root -> right);
}
};

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