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>

Author: ChunelFeng

lcof/面试题28. 对称的二叉树/README.md

@@ -150,6 +150,46 @@ func isSymme(left *TreeNode, right *TreeNode) bool {
 }
 ```
 
+### **C++**
+
+```cpp
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+ * };
+ */
+
+class Solution {
+public:
+    bool isSymmetric(TreeNode* a, TreeNode* b) {
+        // 均为空，则直接返回true。有且仅有一个不为空，则返回false
+        if (a == nullptr && b == nullptr) {
+            return true;
+        } else if (a == nullptr && b != nullptr) {
+            return false;
+        } else if (a != nullptr && b == nullptr) {
+            return false;
+        }
+
+        // 判定值是否相等，和下面的节点是否对称
+        return (a->val == b->val) && isSymmetric(a->left, b->right) && isSymmetric(a->right, b->left);
+    }
+
+    bool isSymmetric(TreeNode* root) {
+        if (root == nullptr) {
+            return true;
+        }
+
+        return isSymmetric(root->left, root->right);
+    }
+};
+
+```
+
 ### **...**
 
 ```

lcof/面试题28. 对称的二叉树/Solution.cpp

@@ -0,0 +1,34 @@
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+ * };
+ */
+
+class Solution {
+public:
+    bool isSymmetric(TreeNode* a, TreeNode* b) {
+        // 均为空，则直接返回true。有且仅有一个不为空，则返回false
+        if (a == nullptr && b == nullptr) {
+            return true;
+        } else if (a == nullptr && b != nullptr) {
+            return false;
+        } else if (a != nullptr && b == nullptr) {
+            return false;
+        }
+
+        // 判定值是否相等，和下面的节点是否对称
+        return (a->val == b->val) && isSymmetric(a->left, b->right) && isSymmetric(a->right, b->left);
+    }
+
+    bool isSymmetric(TreeNode* root) {
+        if (root == nullptr) {
+            return true;
+        }
+
+        return isSymmetric(root->left, root->right);
+    }
+};