From 7d8e9e8fae9d580013b717ff882bd9a942fe0e4f Mon Sep 17 00:00:00 2001 From: Guang-Hou <87743934+Guang-Hou@users.noreply.github.com> Date: Wed, 9 Feb 2022 16:37:44 -0500 Subject: [PATCH] =?UTF-8?q?Update=200701.=E4=BA=8C=E5=8F=89=E6=90=9C?= =?UTF-8?q?=E7=B4=A2=E6=A0=91=E4=B8=AD=E7=9A=84=E6=8F=92=E5=85=A5=E6=93=8D?= =?UTF-8?q?=E4=BD=9C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit Python 递归法- 无返回值的一种更简单的实现。 --- .../0701.二叉搜索树中的插入操作.md | 20 +++++++++++++++++++ 1 file changed, 20 insertions(+) diff --git a/problems/0701.二叉搜索树中的插入操作.md b/problems/0701.二叉搜索树中的插入操作.md index 468f2675..5e9fbdfe 100644 --- a/problems/0701.二叉搜索树中的插入操作.md +++ b/problems/0701.二叉搜索树中的插入操作.md @@ -310,6 +310,26 @@ class Solution: return root ``` +**递归法** - 无返回值 - another easier way +```python +class Solution: + def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]: + newNode = TreeNode(val) + if not root: return newNode + + if not root.left and val < root.val: + root.left = newNode + if not root.right and val > root.val: + root.right = newNode + + if val < root.val: + self.insertIntoBST(root.left, val) + if val > root.val: + self.insertIntoBST(root.right, val) + + return root +``` + **迭代法** 与无返回值的递归函数的思路大体一致 ```python