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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.

判断一个二叉树是不是自己的镜像