Commit d635a4c

ChunelFengjunfeng.fj

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junfeng.fj

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更新了[lcof-28.对称的二叉树]的cpp解法 (doocs#349)

Co-authored-by: junfeng.fj <junfeng.fj@alibaba-inc.com>

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Lines changed: 40 additions & 0 deletions

Original file line number	Diff line number	Diff line change
`@@ -150,6 +150,46 @@ func isSymme(left TreeNode, right TreeNode) bool {`
`150`	`150`	`}`
`151`	`151`	```
`152`	`152`
	`153`	`+### C++`
	`154`	`+`
	`155`	+```cpp
	`156`	`+/**`
	`157`	`+ * Definition for a binary tree node.`
	`158`	`+ * struct TreeNode {`
	`159`	`+ * int val;`
	`160`	`+ * TreeNode *left;`
	`161`	`+ * TreeNode *right;`
	`162`	`+ * TreeNode(int x) : val(x), left(NULL), right(NULL) {}`
	`163`	`+ * };`
	`164`	`+ */`
	`165`	`+`
	`166`	`+class Solution {`
	`167`	`+public:`
	`168`	`+ bool isSymmetric(TreeNode* a, TreeNode* b) {`
	`169`	`+ // 均为空,则直接返回true。有且仅有一个不为空,则返回false`
	`170`	`+ if (a == nullptr && b == nullptr) {`
	`171`	`+ return true;`
	`172`	`+ } else if (a == nullptr && b != nullptr) {`
	`173`	`+ return false;`
	`174`	`+ } else if (a != nullptr && b == nullptr) {`
	`175`	`+ return false;`
	`176`	`+ }`
	`177`	`+`
	`178`	`+ // 判定值是否相等,和下面的节点是否对称`
	`179`	`+ return (a->val == b->val) && isSymmetric(a->left, b->right) && isSymmetric(a->right, b->left);`
	`180`	`+ }`
	`181`	`+`
	`182`	`+ bool isSymmetric(TreeNode* root) {`
	`183`	`+ if (root == nullptr) {`
	`184`	`+ return true;`
	`185`	`+ }`
	`186`	`+`
	`187`	`+ return isSymmetric(root->left, root->right);`
	`188`	`+ }`
	`189`	`+};`
	`190`	`+`
	`191`	+```
	`192`	`+`
`153`	`193`	`### ...`
`154`	`194`
`155`	`195`	```

Lines changed: 34 additions & 0 deletions

Original file line number	Diff line number	Diff line change
`@@ -0,0 +1,34 @@`
	`1`	`+/**`
	`2`	`+ * Definition for a binary tree node.`
	`3`	`+ * struct TreeNode {`
	`4`	`+ * int val;`
	`5`	`+ * TreeNode *left;`
	`6`	`+ * TreeNode *right;`
	`7`	`+ * TreeNode(int x) : val(x), left(NULL), right(NULL) {}`
	`8`	`+ * };`
	`9`	`+ */`
	`10`	`+`
	`11`	`+class Solution {`
	`12`	`+public:`
	`13`	`+ bool isSymmetric(TreeNode* a, TreeNode* b) {`
	`14`	`+ // 均为空,则直接返回true。有且仅有一个不为空,则返回false`
	`15`	`+ if (a == nullptr && b == nullptr) {`
	`16`	`+ return true;`
	`17`	`+ } else if (a == nullptr && b != nullptr) {`
	`18`	`+ return false;`
	`19`	`+ } else if (a != nullptr && b == nullptr) {`
	`20`	`+ return false;`
	`21`	`+ }`
	`22`	`+`
	`23`	`+ // 判定值是否相等,和下面的节点是否对称`
	`24`	`+ return (a->val == b->val) && isSymmetric(a->left, b->right) && isSymmetric(a->right, b->left);`
	`25`	`+ }`
	`26`	`+`
	`27`	`+ bool isSymmetric(TreeNode* root) {`
	`28`	`+ if (root == nullptr) {`
	`29`	`+ return true;`
	`30`	`+ }`
	`31`	`+`
	`32`	`+ return isSymmetric(root->left, root->right);`
	`33`	`+ }`
	`34`	`+};`

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