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) -
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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) -
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