Determining whether a binary tree is a mirror symmetric

Given a binary tree, check if it is mirror-symmetrical. 

 For example, a 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 a mirror image: 

     1 
    / \
    2    2 
    \ \ 
   . 3     . 3

Thinking: binary tree with a height greater than 2 bar for example, which is the first example above, the left child node and the node as long as the right child of 1 is equal to 2, the left and right child nodes and the node point 2 of the children are equal, we believe that is a mirror, provided that nodes 1 and 2 sibling node;

Recursive achieve the following:

public static boolean isSymmetric(TreeNode root) {
        if(root == null) return true;
        return in(root.left,root.right);
    }


    //递归
    public static boolean in(TreeNode l1,TreeNode l2){
        if(l1 == null && l2 == null) return true;
        if(l1 == null || l2 == null) return false;
        return l1.val == l2.val && in(l1.left,l2.right) && in(l1.right,l2.left);
    }

Non-recursive as follows:

//迭代
    public static boolean isSymm(TreeNode root){
        if(root == null) return true;
        Queue<TreeNode> q = new LinkedList<>();
        q.add(root);
        q.add(root);
        while (!q.isEmpty()) {
            TreeNode t1 = q.poll();
            TreeNode t2 = q.poll();
            if (t1 == null && t2 == null) continue;
            if (t1 == null || t2 == null) return false;
            if (t1.val != t2.val) return false;
            q.add(t1.left);
            q.add(t2.right);
            q.add(t1.right);
            q.add(t2.left);
        }
        return true;
    }