判断一个二叉树是不是自己的镜像

bool helper(TreeNode *pA, TreeNode *pB) {
        if (!pA && !pB) return true;
        if (!pA || !pB) return false;  // only one has node in a tree and b tree
        if (pA->val != pB->val) return false;
        return helper(pA->left, pB->.right) && helper(pA->right, pB->left); // a goes in in-order traversal, b goes right first then left.
    }
    
public:
    bool isSymmetric(TreeNode *root) {
        if (!root)
            return true;
        
        return helper(root->left, root->right);
    }

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.