🐱(link): 24. 两两交换链表中的节点

补充递归题解

Author: JalanJiang

docs/data-structure/linked_list/README.md

@@ -166,14 +166,12 @@ class Solution(object):
 
 ## 24. 两两交换链表中的节点
 
-### 思路
+### 解一：迭代
 
 - 为方便统一化创建空节点 `cur`
 - 交换呗，没啥好说
 - 注意返回头节点
 
-### Python
-
 ```python
 class Solution(object):
  def swapPairs(self, head):
@@ -195,6 +193,56 @@ class Solution(object):
  return first.next
 ```
 
+### 解二：递归
+
+```python
+# Definition for singly-linked list.
+# class ListNode:
+# def __init__(self, x):
+# self.val = x
+# self.next = None
+
+class Solution:
+ def swapPairs(self, head: ListNode) -> ListNode:
+ def helper(node, pre):
+ if node is None or node.next is None:
+ return 
+ next_node = node.next
+ # 指针交换
+ node.next = next_node.next
+ next_node.next = node
+ pre.next = next_node
+ helper(node.next, node)
+ 
+ ans = ListNode(0)
+ ans.next = head
+ helper(head, ans)
+ return ans.next
+```
+
+优雅递归：
+
+```python
+# Definition for singly-linked list.
+# class ListNode:
+# def __init__(self, x):
+# self.val = x
+# self.next = None
+
+class Solution:
+ def swapPairs(self, head: ListNode) -> ListNode:
+ if head is None or head.next is None:
+ return head
+
+ first_node = head
+ second_node = head.next
+
+ first_node.next = self.swapPairs(second_node.next)
+ second_node.next = first_node
+
+ return second_node
+```
+
 ## 25. K 个一组翻转链表
 
 [原题链接](https://leetcode-cn.com/problems/reverse-nodes-in-k-group/)