Symmetric Tree

题目：
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
做这道题，最好的解法我觉得应该是递归求解。开始我想的是先层序遍历，然后判断回文，发现不好使。后来收大神启发，用了递归，效率杠杠的。话不多说，上代码：

    public boolean isSymmetric(TreeNode root) {
        return root == null || helpSmmetric(root.left, root.right);
    }

    public boolean helpSmmetric(TreeNode left, TreeNode right) {
        //递归结束条件节点为空的情况，为空说明至少一个分支已经到低了，后面的也不需要在比较
        if (left == null || right == null) {
            return left == right;
        }
        //递归结束条件节点不为空的情况，只要不等直接返回false，后面不需要比较
        if (left.val != right.val)
            return false;
        //递归调用可确保每次对比的节点都是对称的节点
        return helpSmmetric(left.left, right.right) && helpSmmetric(left.right, right.left);
    }

感觉递归好好用，我开始没用递归，写了好长的代码才勉强实现功能，然而递归核心步骤才几行。