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.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Pair{
    TreeNode x;
    TreeNode y;
    public Pair(TreeNode x, TreeNode y){
        this.x = x;
        this.y = y;
    }
}
public class Solution {
    // recursive
    // public boolean isSymmetric(TreeNode root) {
    //     if(root == null)
    //         return true;
        
    //     return Symmetric(root.left, root.right);
    // }
    // public boolean Symmetric(TreeNode left, TreeNode right){
    //     if(left == null || right == null){
    //         return left == right;
    //     }
    //     if(left.val != right.val){
    //         return false;
    //     }
    //     return Symmetric(left.left, right.right) &&  Symmetric(left.right, right.left);
    // }
    
    public boolean isSymmetric(TreeNode root) {
        if(root == null || root.left == null && root.right == null)
            return true;
        Queue<Pair> queue = new LinkedList<Pair>();
        queue.offer(new Pair(root.left, root.right));
        while(!queue.isEmpty()){
            Pair values= queue.poll();
            if(values.x == null && values.y == null)
                continue;
            if(values.x == null || values.y == null || values.x.val != values.y.val)
               return false;
               
            if(values.x != null && values.y != null){
                queue.offer(new Pair(values.x.left, values.y.right));
                queue.offer(new Pair(values.x.right, values.y.left));
            }
        }
        return true;
    }
}