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更新 0968.监控二叉树.md

CPP代码未语法高亮

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`‎problems/0968.监控二叉树.md`

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Original file line number	Diff line number	Diff line change
`@@ -71,7 +71,7 @@`
`71`	`71`
`72`	`72`	`后序遍历代码如下:`
`73`	`73`
`74`		-```
	`74`	+```CPP
`75`	`75`	`int traversal(TreeNode* cur) {`
`76`	`76`
`77`	`77`	`// 空节点,该节点有覆盖`
`@@ -124,7 +124,7 @@ int traversal(TreeNode* cur) {`
`124`	`124`
`125`	`125`	`代码如下:`
`126`	`126`
`127`		-```
	`127`	+```CPP
`128`	`128`	`// 空节点,该节点有覆盖`
`129`	`129`	`if (cur == NULL) return 2;`
`130`	`130`	```
`@@ -143,7 +143,7 @@ if (cur == NULL) return 2;`
`143`	`143`
`144`	`144`	`代码如下:`
`145`	`145`
`146`		-```
	`146`	+```CPP
`147`	`147`	`// 左右节点都有覆盖`
`148`	`148`	`if (left == 2 && right == 2) return 0;`
`149`	`149`	```
`@@ -163,7 +163,7 @@ left == 2 && right == 0 左节点覆盖,右节点无覆盖`
`163`	`163`	`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。`
`164`	`164`
`165`	`165`	`代码如下:`
`166`		-```
	`166`	+```CPP
`167`	`167`	`if (left == 0 \|\| right == 0) {`
`168`	`168`	`result++;`
`169`	`169`	`return 1;`
`@@ -180,7 +180,7 @@ left == 1 && right == 1 左右节点都有摄像头`
`180`	`180`
`181`	`181`	`代码如下:`
`182`	`182`
`183`		-```
	`183`	+```CPP
`184`	`184`	`if (left == 1 \|\| right == 1) return 2;`
`185`	`185`	```
`186`	`186`
`@@ -198,7 +198,7 @@ if (left == 1 \|\| right == 1) return 2;`
`198`	`198`
`199`	`199`	`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:`
`200`	`200`
`201`		-```
	`201`	+```CPP
`202`	`202`	`int minCameraCover(TreeNode* root) {`
`203`	`203`	`result = 0;`
`204`	`204`	`if (traversal(root) == 0) { // root 无覆盖`

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