101 Symmetric Tree

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.

Solution (recursive):

/**
 * 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 isMirror(TreeNode* root1, TreeNode* root2) {
        if ( root1 == NULL ) return (root2 == NULL );
        if ( root2 == NULL ) return false;
        if ( root1->val != root2->val ) return false;
        if ( not isMirror(root1->left, root2->right) ) return false;
        if ( not isMirror(root1->right, root2->left) ) return false;
        return true;
    }
    bool isSymmetric(TreeNode* root) {
        if ( root == NULL ) return true;
        return (isMirror(root->left, root->right));
    }
};

Solution (iterative):

/**
 * 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 isSymmetric(TreeNode* root) {
        if ( root == NULL ) return true;
        // use level-ordere traversal result
        vector<TreeNode*> res;
        res.push_back(root->left);
        res.push_back(root->right);
        while ( res.size() != 0 ) {
            int nres = res.size();
            int i = 0, j = nres-1;
            while ( i <= j-1 ) {
                if ( res[i] == NULL and res[j] != NULL ) return false;
                if ( res[i] != NULL and res[j] == NULL ) return false;
                if ( res[i] == NULL and res[j] == NULL ) { i += 1; j -= 1; continue;}
                if ( res[i]->val != res[j]->val ) return false;
                i += 1;
                j -= 1;
            }
            vector<TreeNode*> res2;
            for ( i = 0; i <= nres-1; i++ ) {
                if ( res[i] == NULL ) continue;
                res2.push_back(res[i]->left);
                res2.push_back(res[i]->right);
            }
            res = res2;
        }
        return true;
    }
};