leetcode 13: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.

/**
 * 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 isSymRec(TreeNode *left, TreeNode * right){
        if(left == NULL && right==NULL){
            return true;
        } else if(left == NULL || right == NULL) {
            return false;
        }
        
        return ( left->val == right->val && isSymRec( left->left, right->right) && 
            isSymRec( left->right, right->left) );
        //both are not NULL.       
    }

    bool isSymmetric(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(root == NULL) return true;
        return isSymRec( root->left, root->right);
    }
};

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isSymmetric(TreeNode root) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(root==null) {
            return true;
        }
        
        return isSysRec(root.left, root.right);
    }
    
    private boolean isSysRec(TreeNode left, TreeNode right){
        if(left==null && right==null) {
            return true;
        } else if( left==null || right==null) {
            return false;
        }
        
        return (left.val==right.val) && isSysRec(left.left, right.right) && isSysRec(left.right, right.left);
    }
}