更新了[lcof-68.2 二叉树的最近公共祖先]

lcof/面试题57 - II. 和为s的连续正数序列/README.md

@@ -113,6 +113,55 @@ var findContinuousSequence = function (target) {
 };
 ```
 
+### **C++**
+
+```cpp
+class Solution {
+public:
+    vector<int> build(int small, int big) {
+        vector<int> ret;
+        for (int i = small; i <= big; i++) {
+            ret.push_back(i);
+        }
+
+        return ret;
+    }
+
+    vector<vector<int>> findContinuousSequence(int target) {
+        vector<vector<int>> ret;
+        int small = 1;
+        int big = 2;
+        int mid = (target + 1) / 2;
+        int curSum = small + big;
+
+        if (target < 3) {
+            ret;
+        }
+
+        while(small < mid) {
+            if (curSum == target) {
+                ret.push_back(build(small, big));
+            }
+
+            while (curSum > target && small < mid) {
+                // 一直减去，减去到比target小停止
+                curSum -= small;
+                small++;
+
+                if (curSum == target && small < mid) {
+                    ret.push_back(build(small, big));
+                }
+            }
+
+            big++;
+            curSum += big;
+        }
+
+        return ret;
+    }
+};
+```
+
 ### **...**
 
 ```

lcof/面试题57 - II. 和为s的连续正数序列/Solution.cpp

@@ -0,0 +1,44 @@
+class Solution {
+public:
+    vector<int> build(int small, int big) {
+        vector<int> ret;
+        for (int i = small; i <= big; i++) {
+            ret.push_back(i);
+        }
+
+        return ret;
+    }
+
+    vector<vector<int>> findContinuousSequence(int target) {
+        vector<vector<int>> ret;
+        int small = 1;
+        int big = 2;
+        int mid = (target + 1) / 2;
+        int curSum = small + big;
+
+        if (target < 3) {
+            ret;
+        }
+
+        while(small < mid) {
+            if (curSum == target) {
+                ret.push_back(build(small, big));
+            }
+
+            while (curSum > target && small < mid) {
+                // 一直减去，减去到比target小停止
+                curSum -= small;
+                small++;
+
+                if (curSum == target && small < mid) {
+                    ret.push_back(build(small, big));
+                }
+            }
+
+            big++;
+            curSum += big;
+        }
+
+        return ret;
+    }
+};

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;
+        }
+    }
+};