From 65fae3b97541e4e73d8a9c1bd7b0483d8b49b359 Mon Sep 17 00:00:00 2001 From: zhicheng lee <904688436@qq.com> Date: Mon, 5 Sep 2022 16:20:42 +0800 Subject: [PATCH] =?UTF-8?q?=E6=9B=B4=E6=96=B0=200968.=E7=9B=91=E6=8E=A7?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit CPP代码未语法高亮 --- problems/0968.监控二叉树.md | 12 ++++++------ 1 file changed, 6 insertions(+), 6 deletions(-) diff --git a/problems/0968.监控二叉树.md b/problems/0968.监控二叉树.md index b17ff080..717112a7 100644 --- a/problems/0968.监控二叉树.md +++ b/problems/0968.监控二叉树.md @@ -71,7 +71,7 @@ 后序遍历代码如下: -``` +```CPP int traversal(TreeNode* cur) { // 空节点,该节点有覆盖 @@ -124,7 +124,7 @@ int traversal(TreeNode* cur) { 代码如下: -``` +```CPP // 空节点,该节点有覆盖 if (cur == NULL) return 2; ``` @@ -143,7 +143,7 @@ if (cur == NULL) return 2; 代码如下: -``` +```CPP // 左右节点都有覆盖 if (left == 2 && right == 2) return 0; ``` @@ -163,7 +163,7 @@ left == 2 && right == 0 左节点覆盖,右节点无覆盖 此时摄像头的数量要加一,并且return 1,代表中间节点放摄像头。 代码如下: -``` +```CPP if (left == 0 || right == 0) { result++; return 1; @@ -180,7 +180,7 @@ left == 1 && right == 1 左右节点都有摄像头 代码如下: -``` +```CPP if (left == 1 || right == 1) return 2; ``` @@ -198,7 +198,7 @@ if (left == 1 || right == 1) return 2; 所以递归结束之后,还要判断根节点,如果没有覆盖,result++,代码如下: -``` +```CPP int minCameraCover(TreeNode* root) { result = 0; if (traversal(root) == 0) { // root 无覆盖