Easy
Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).
Example 1:

Input: root = [1,2,2,3,4,4,3]
Output: true
Example 2:

Input: root = [1,2,2,null,3,null,3]
Output: false
Constraints:
[1, 1000].-100 <= Node.val <= 100Follow up: Could you solve it both recursively and iteratively?
using LeetCodeNet.Com_github_leetcode;
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public bool IsSymmetric(TreeNode root) {
        if (root == null) {
            return true;
        }
        return Helper(root.left, root.right);
    }
    private bool Helper(TreeNode leftNode, TreeNode rightNode) {
        if (leftNode == null || rightNode == null) {
            return leftNode == null && rightNode == null;
        }
        if (leftNode.val != rightNode.val) {
            return false;
        }
        return Helper(leftNode.left, rightNode.right) && Helper(leftNode.right, rightNode.left);
    }
}