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.

Solution1： recursion

1. Bottom-up,
   看left.left && right.right 和 left.right && right.left是否是symmetric, 如果是，再看left.val == right.val?
2. Base Case：
   left 和right任一一个不为null，则不是对称，返回false
   left和right都为null，则对称，返回true

class Solution {
    //// ********* Recursion solution ****************/
    public boolean isSymmetric(TreeNode root) {
        if (root == null) {
            return true;
        }
        
        return isSymmetricHelper (root.left, root.right);
    }
    
    private boolean isSymmetricHelper (TreeNode left, TreeNode right) {
        if (left == null && right == null) {
            return true;
        }
        
        if (left == null || right == null) {
            return false;
        }
        
        return left.val == right.val && isSymmetricHelper (left.left, right.right) && isSymmetricHelper (left.right, right.left);
    }
    

Solution2： Iteration

1. 与Same Tree那题很像，比较就是root.left, root.right是否的对称的same tree

//// ********* Iteration solution ****************/
    public boolean isSymmetric(TreeNode root) {
        if (root == null) {
            return true;
        }

        Queue<TreeNode> tracker1 = new LinkedList<> ();
        Queue<TreeNode> tracker2 = new LinkedList<> ();
        
        tracker1.offer (root.left);
        tracker2.offer (root.right);
        
        while (!tracker1.isEmpty () && !tracker2.isEmpty ()) {
            TreeNode left = tracker1.poll ();
            TreeNode right = tracker2.poll ();
            
            // 有可能某一层，不是满的
            if (left == null && right == null) {
                continue;
            }
            
            if (left == null || right == null) {
                return false;
            }
            
            if (left.val != right.val) {
                return false;
            }
            
            tracker1.offer (left.left);
            tracker1.offer (left.right);
            
            tracker2.offer (right.right);
            tracker2.offer (right.left);
        }
        
        if (!tracker1.isEmpty () || !tracker2.isEmpty ()) {
            return false;
        }
        
        return true;
    }
    //// ********* Iteration solution END ****************/