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feat:azl397985856#92 增加注释 #236增加扩展 backlog增加题目

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	`1`	`+/*`
	`2`	`+ *@lc app=leetcode id=538 lang=javascript`
	`3`	`+ *`
	`4`	`+ * [538] Convert BST to Greater Tree`
	`5`	`+ *`
	`6`	`+ * https://leetcode.com/problems/convert-bst-to-greater-tree/description/`
	`7`	`+ *`
	`8`	`+ * algorithms`
	`9`	`+ * Easy (50.04%)`
	`10`	`+ * Total Accepted: 75.4K`
	`11`	`+ * Total Submissions: 149K`
	`12`	`+ * Testcase Example: '[5,2,13]'`
	`13`	`+ *`
	`14`	`+ * Given a Binary Search Tree (BST), convert it to a Greater Tree such that`
	`15`	`+ * every key of the original BST is changed to the original key plus sum of all`
	`16`	`+ * keys greater than the original key in BST.`
	`17`	`+ *`
	`18`	`+ *`
	`19`	`+ * Example:`
	`20`	`+ *`
	`21`	`+ * Input: The root of a Binary Search Tree like this:`
	`22`	`+ * ⁠ 5`
	`23`	`+ * ⁠ / \`
	`24`	`+ * ⁠ 2 13`
	`25`	`+ *`
	`26`	`+ * Output: The root of a Greater Tree like this:`
	`27`	`+ * ⁠ 18`
	`28`	`+ * ⁠ / \`
	`29`	`+ * ⁠ 20 13`
	`30`	`+ *`
	`31`	`+ *`
	`32`	`+ */`
	`33`	`+/**`
	`34`	`+ * Definition for a binary tree node.`
	`35`	`+ * function TreeNode(val) {`
	`36`	`+ * this.val = val;`
	`37`	`+ * this.left = this.right = null;`
	`38`	`+ * }`
	`39`	`+ */`
	`40`	`+/**`
	`41`	`+ *@param {TreeNode} root`
	`42`	`+ *@return {TreeNode}`
	`43`	`+ */`
	`44`	`+varconvertBST=function(root){`
	`45`	`+letres=0;`
	`46`	`+functionr(root){`
	`47`	`+if(root===null)returnnull;`
	`48`	`+`
	`49`	`+r(root.right);`
	`50`	`+`
	`51`	`+root.val+=res;`
	`52`	`+`
	`53`	`+res=+root.val;`
	`54`	`+`
	`55`	`+r(root.left);`
	`56`	`+`
	`57`	`+returnroot;`
	`58`	`+}`
	`59`	`+r(root);`
	`60`	`+returnroot;`
	`61`	`+};`

Lines changed: 3 additions & 0 deletions

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`@@ -147,3 +147,6 @@ var lowestCommonAncestor = function(root, p, q) {`
`147`	`147`	`};`
`148`	`148`	```
`149`	`149`
	`150`	`+##扩展`
	`151`	+如果递归的结束条件改为`if (!root \|\| root.left === p \|\| root.right === q) return root;` 代表的是嗯呢么意思，对结果有什么样的影响？
	`152`	`+`

Lines changed: 7 additions & 5 deletions

Original file line number	Diff line number	Diff line change
`@@ -89,14 +89,16 @@ Output: 1->4->3->2->5->NULL`
`89`	`89`	`*@return{ListNode}`
`90`	`90`	`*/`
`91`	`91`	`varreverseBetween=function(head,m,n) {`
	`92`	`+// 虚拟节点，简化操作`
`92`	`93`	`constdummyHead= {`
`93`	`94`	`next: head`
`94`	`95`	`}`
`95`	`96`
`96`		`-let current=dummyHead.next;`
`97`		`-let pre= current;`
`98`		`-let index=0;`
	`97`	`+let current=dummyHead.next;// 当前遍历的节点`
	`98`	`+let pre= current;// 因为要反转，因此我们需要记住前一个节点`
	`99`	`+let index=0;// 链表索引，用来判断是否是特殊位置（头尾位置）`
`99`	`100`
	`101`	`+// 上面提到的四个特殊节点`
`100`	`102`	`let midStartNode=null;`
`101`	`103`	`let preMidStartNode=null;`
`102`	`104`	`let midEndNode=null;`
`@@ -106,11 +108,12 @@ var reverseBetween = function(head, m, n) {`
`106`	`108`	`constnext=current.next;`
`107`	`109`	`index++;`
`108`	`110`
`109`		`-//reverse linked list`
	`111`	`+//对 (m - n) 范围内的节点进行反转`
`110`	`112`	`if (index> m&& index<= n) {`
`111`	`113`	`current.next= pre;`
`112`	`114`	`}`
`113`	`115`
	`116`	`+// 下面四个if都是边界, 用于更新四个特殊节点的值`
`114`	`117`	`if (index=== m-1) {`
`115`	`118`	`preMidStartNode= current;`
`116`	`119`	`}`
`@@ -120,7 +123,6 @@ var reverseBetween = function(head, m, n) {`
`120`	`123`
`121`	`124`	`if (index=== n+1) {`
`122`	`125`	`postMidEndNode= current;`
`123`		`-postMidEndNode.next`
`124`	`126`	`}`
`125`	`127`
`126`	`128`	`if (index=== n) {`

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