//101

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.

class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        if(root==NULL){
            return true;
        }

        return isMirrow(root->left,root->right);

    }
private:
//aux functions
    bool isMirrow(TreeNode* p,TreeNode*q){
        if(p==NULL && q==NULL){
            return true;
        }

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

        if(p->val==q->val){
            bool left=isMirrow(p->left,q->right);
            bool right=isMirrow(p->right,q->left);

            return left && right;
        } else{
            return false;
        }
    }
};