Symmetric Tree

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.

思路：这个没啥可说的。两种解法递归和迭代。

public boolean isSymmetric(TreeNode root) {
        if (root == null) return true;
        return helper(root.left, root.right);
    }
    private boolean helper(TreeNode left, TreeNode right) {
        if (left == null) return right == null;
        if (right == null) return left == null;
        if (left.val == right.val) {
            return helper(left.left, right.right) && helper(left.right, right.left);
        }else{
            return false;
        }
    }

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