Symmetric Tree

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.

// ITERATIVE VERSION:

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSymmetric(TreeNode *root) {
        
        // use BFS, using two queue
        queue<TreeNode *> p;
        queue<TreeNode *> q;
        p.push(root);
        q.push(root);
        TreeNode * pnode, * qnode;
        while (!p.empty() && !q.empty()){
            pnode = p.front(); p.pop();
            qnode = q.front(); q.pop();
            if (pnode==NULL && qnode==NULL)
                continue;
            else if (!(pnode && qnode) || (pnode->val != qnode->val)) return false;
            else {
                p.push(pnode->left); p.push(pnode->right);
                q.push(qnode->right); q.push(qnode->left);
            }
        }
        return true;
    }
};

//==============================================================

// RECURSIVE VERSION:

class Solution {
public:
    bool status;
    void helper(TreeNode * p, TreeNode * q){
        if (!status || (p==NULL && q==NULL))
            return;
        if (!(p&&q) || (p->val != q->val)){
            status = false;
            return;
        }
        else{
            helper(p->left, q->right);
            helper(p->right, q->left);
        }
    }
    bool isSymmetric(TreeNode *root) {
        status = true;
        helper(root, root);
        return status;
    }
};