From ad538d8e48e21d5c0910e47912c6c3213379ac15 Mon Sep 17 00:00:00 2001 From: youngyangyang04 <826123027@qq.com> Date: Sun, 16 May 2021 20:06:05 +0800 Subject: [PATCH] =?UTF-8?q?=E4=BF=AE=E6=AD=A3Markdown=E4=BB=A3=E7=A0=81?= =?UTF-8?q?=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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