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.

For this problem I want to record the iterative solution.

Recusive:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root==null) return true;
        return helper(root.left, root.right);
    }

    public boolean helper(TreeNode lt, TreeNode rt) {
        if(lt==null||rt==null) return (rt==lt);
        if(lt.val==rt.val) return helper(lt.left, rt.right)&&helper(lt.right, rt.left);
        return false;
    }
}

Iterative:

The below solution is BFS and iterative.
In Queue (LinkedList) , we use Queue.offer(element) to insert (will not throw Null exception) and Queue.poll() to pop the top element.

public class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null || root.left == null && root.right == null){
            return true;
        }
        if (root.left == null || root.right == null){
            return false;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root.left);
        queue.offer(root.right);
        while (!queue.isEmpty()) {
            TreeNode left = queue.poll();
            TreeNode right = queue.poll();
            if (left.val != right.val){
                return false;
            }

            if (left.left != null && right.right != null){
                queue.offer(left.left);
                queue.offer(right.right);
            }
            else if (left.left == null && right.right != null || right.right == null && left.left != null){
                return false;
            }
            if (left.right != null && right.left != null){
                queue.offer(left.right);
                queue.offer(right.left);
            }
            else if (left.right == null && right.left != null || right.left == null && left.right != null){
                return false;
            }
        }
        return true;
    }
}