From 92c6be6e0a60460314b543d9ba327f9a79493c60 Mon Sep 17 00:00:00 2001
From: QuinnDK <39618652+QuinnDK@users.noreply.github.com>
Date: Sun, 16 May 2021 19:55:03 +0800
Subject: [PATCH 1/2] =?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
---
.../0701.二叉搜索树中的插入操作.md | 16 +++++++++++++++-
1 file changed, 15 insertions(+), 1 deletion(-)
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 ee690d04..122a1d2a 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
@@ -271,6 +271,20 @@ class Solution:
Go：
+```Go
+func insertIntoBST(root *TreeNode, val int) *TreeNode {
+ if root == nil {
+ root = &TreeNode{Val: val}
+ return root
+ }
+ if root.Val > val {
+ root.Left = insertIntoBST(root.Left, val)
+ } else {
+ root.Right = insertIntoBST(root.Right, val)
+ }
+ return root
+}
+```
@@ -279,4 +293,4 @@ Go：
* 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
* B站视频：[代码随想录](https://space.bilibili.com/525438321)
* 知识星球：[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
\ No newline at end of file
+
From 21a4b9d64b640e233d6c41fe732693e4075e5236 Mon Sep 17 00:00:00 2001
From: youngyangyang04 <826123027@qq.com>
Date: Sun, 16 May 2021 20:06:05 +0800
Subject: [PATCH 2/2] =?UTF-8?q?=E4=BF=AE=E6=AD=A3Markdown=E4=BB=A3?=
=?UTF-8?q?=E7=A0=81=E5=9D=97=E8=AF=AD=E6=B3=95?=
MIME-Version: 1.0
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: 8bit
---
problems/0257.二叉树的所有路径.md | 8 ++++----
1 file changed, 4 insertions(+), 4 deletions(-)
diff --git a/problems/0257.二叉树的所有路径.md b/problems/0257.二叉树的所有路径.md
index 73387257..29bcdf41 100644
--- a/problems/0257.二叉树的所有路径.md
+++ b/problems/0257.二叉树的所有路径.md
@@ -77,7 +77,7 @@ if (cur->left == NULL && cur->right == NULL) {
这里我们先使用vector结构的path容器来记录路径，那么终止处理逻辑如下：
-```
+```C++
if (cur->left == NULL && cur->right == NULL) { // 遇到叶子节点
string sPath;
for (int i = 0; i < path.size() - 1; i++) { // 将path里记录的路径转为string格式
@@ -113,7 +113,7 @@ if (cur->right) {
那么回溯要怎么回溯呢，一些同学会这么写，如下：
-```
+```C++
if (cur->left) {
traversal(cur->left, path, result);
}
@@ -129,7 +129,7 @@ path.pop_back();
那么代码应该这么写：
-```
+```C++
if (cur->left) {
traversal(cur->left, path, result);
path.pop_back(); // 回溯
@@ -335,4 +335,4 @@ Go：
* 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
* B站视频：[代码随想录](https://space.bilibili.com/525438321)
* 知识星球：[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
-
\ No newline at end of file
+
