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/**
 * 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
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
 
 public class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root == null)
            return true;
        return isSymmetric(root.left, root.right);
    }
    public boolean isSymmetric(TreeNode left, TreeNode right) {
        if(left == null && right == null) 
            return true;
        if(left == null || right == null) 
            return false;
        if(left.val != right.val) 
            return false;
        return isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left);
 }