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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.
递归：

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

迭代：

public boolean isSymmetric(TreeNode root) {
        if(root == null) return true;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        stack.push(root.left);
        stack.push(root.right);
        while(!stack.empty()) {
            TreeNode right = stack.pop();
            TreeNode left = stack.pop();
            if (left == null && right == null) continue;
            else if (left == null || right == null || right.val != left.val) return false;
            stack.push(left.left);
            stack.push(right.right);
            stack.push(left.right);
            stack.push(right.left);
        }
        return true;
    }