更新了[lcof-68.2 二叉树的最近公共祖先] (#348)

* 增加了[lcof-18.删除链表的节点]和[lcof-40.最小的k个数]的cpp解法,基本双优

* 优化[lcof-40.最小的k个数]格式

* 更新了[lcof-57.和为s的连续正整数序列]的cpp版本解法

* 更新了[lcof-68.2 二叉树的最近公共祖先]

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

Author: ChunelFeng

lcof/面试题68 - II. 二叉树的最近公共祖先/README.md

@@ -129,6 +129,44 @@ var lowestCommonAncestor = function (root, p, q) {
 };
 ```
 
+### **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:
+    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
+        // 如果找到val，层层向上传递该root
+        if (nullptr == root || p->val == root->val || q->val == root->val) {
+            return root;
+        }
+
+        TreeNode* left = lowestCommonAncestor(root->left, p, q);
+        TreeNode* right = lowestCommonAncestor(root->right, p, q);
+
+        if (left != nullptr && right != nullptr) {
+            // 如果两边都可以找到
+            return root;
+        } else if (left == nullptr) {
+            // 如果左边没有找到，则直接返回右边内容
+            return right;
+        } else {
+            return left;
+        }
+    }
+};
+
+```
+
 ### **...**
 
 ```

lcof/面试题68 - II. 二叉树的最近公共祖先/Solution.cpp

@@ -0,0 +1,32 @@
+/**
+ * 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:
+    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
+        // 如果找到val，层层向上传递该root
+        if (nullptr == root || p->val == root->val || q->val == root->val) {
+            return root;
+        }
+
+        TreeNode* left = lowestCommonAncestor(root->left, p, q);
+        TreeNode* right = lowestCommonAncestor(root->right, p, q);
+
+        if (left != nullptr && right != nullptr) {
+            // 如果两边都可以找到
+            return root;
+        } else if (left == nullptr) {
+            // 如果左边没有找到，则直接返回右边内容
+            return right;
+        } else {
+            return left;
+        }
+    }
+};