LeetCode 对称二叉树

https://leetcode-cn.com/problems/symmetric-tree/description/

我的解决方案：比较笨拙，我直接按照左后根遍历一遍，然后再按照右后根遍历一遍，最后比较结果

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) { val = x; }
} 

public class Solution {
    public static String lTreeToString(TreeNode tree, String string) {
        if(tree==null) {
            string+=null;
            return string;
        }
        string = lTreeToString(tree.left, string);
        string = lTreeToString(tree.right, string);
        string += Integer.toString(tree.val);
        return string;
    }

    public static String rTreeToString(TreeNode tree, String string) {
        if(tree==null) {
            string+=null;
            return string;
        }
        string = rTreeToString(tree.right, string);
        string = rTreeToString(tree.left, string);
        string += Integer.toString(tree.val);
        return string;
    }
    public static boolean isSymmetric(TreeNode root) { 
        //我的思路是分别对树进行左后根遍历和右后根遍历
        //最后比较形成的结果是否相同
        String lString=lTreeToString(root, "");
        String rString=rTreeToString(root, "");
        return lString.equals(rString)?true:false;
    }
}

提交记录中耗时最短的代码：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root == null) return true;
        return isSym(root.left, root.right);
    }

    private boolean isSym(TreeNode left, TreeNode right) {
        if(left == null|| right == null) {
            return left==right?true:false;
        }
        if(left.val != right.val) return false;
        return isSym(left.left, right.right) && isSym(left.right, right.left);
    }
}

不得不说自己的思维还是不行啊。。。。