Update 144.binary-tree-preorder-traversal.md

problems/144.binary-tree-preorder-traversal.md

@@ -33,7 +33,7 @@ Follow up: Recursive solution is trivial, could you do it iteratively?
 前序遍历是`根左右`的顺序，注意是`根`开始，那么就很简单。直接先将根节点入栈，然后
 看有没有右节点，有则入栈，再看有没有左节点，有则入栈。 然后出栈一个元素，重复即可。
 
-> 其他树的非递归遍历课没这么简单
+> 其他树的非递归遍历可没这么简单
 
 ![](https://tva1.sinaimg.cn/large/007S8ZIlly1gg7v5hztd2j30zu0ntwiu.jpg)
 
@@ -62,44 +62,6 @@ Follow up: Recursive solution is trivial, could you do it iteratively?
 JavaScript Code:
 
 ```js
-/*
- * @lc app=leetcode id=144 lang=javascript
- *
- * [144] Binary Tree Preorder Traversal
- *
- * https://leetcode.com/problems/binary-tree-preorder-traversal/description/
- *
- * algorithms
- * Medium (50.36%)
- * Total Accepted:    314K
- * Total Submissions: 621.2K
- * Testcase Example:  '[1,null,2,3]'
- *
- * Given a binary tree, return the preorder traversal of its nodes' values.
- *
- * Example:
- *
- *
- * Input: [1,null,2,3]
- * ⁠  1
- * ⁠   \
- * ⁠    2
- * ⁠   /
- * ⁠  3
- *
- * Output: [1,2,3]
- *
- *
- * Follow up: Recursive solution is trivial, could you do it iteratively?
- *
- */
-/**
- * Definition for a binary tree node.
- * function TreeNode(val) {
- *     this.val = val;
- *     this.left = this.right = null;
- * }
- */
 /**
  * @param {TreeNode} root
  * @return {number[]}