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?=
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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<int>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)
-<div align="center"><img src=../pics/公众号.png width=450 alt=> </img></div>
\ No newline at end of file
+<div align="center"><img src=../pics/公众号.png width=450 alt=> </img></div>