更新了[25.合并两个排序的链表]和[26.树的子结构]两题的cpp解法

lcof/面试题25. 合并两个排序的链表/README.md

@@ -148,6 +148,34 @@ func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
 }
 ```
 
+### **C++**
+
+```cpp
+class Solution {
+public:
+    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
+        if (nullptr == l1 && nullptr == l2) {
+            return nullptr;    // 两个都为空，则直接返回
+        }
+
+        if (nullptr == l1 || nullptr == l2) {
+            return l1 == nullptr ? l2 : l1;    // 有且仅有一个为空，则返回非空节点
+        }
+
+        ListNode* node = nullptr;
+        if (l1->val > l2->val) {
+            node = l2;
+            node->next = mergeTwoLists(l1, l2->next);
+        } else {
+            node = l1;
+            node->next = mergeTwoLists(l1->next, l2);
+        }
+
+        return node;
+    }
+};
+```
+
 ### **...**
 
 ```

lcof/面试题25. 合并两个排序的链表/Solurion.cpp

@@ -0,0 +1,32 @@
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode(int x) : val(x), next(NULL) {}
+ * };
+ */
+
+class Solution {
+public:
+    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
+        if (nullptr == l1 && nullptr == l2) {
+            return nullptr;
+        }
+
+        if (nullptr == l1 || nullptr == l2) {
+            return l1 == nullptr ? l2 : l1;
+        }
+
+        ListNode* node = nullptr;
+        if (l1->val > l2->val) {
+            node = l2;
+            node->next = mergeTwoLists(l1, l2->next);
+        } else {
+            node = l1;
+            node->next = mergeTwoLists(l1->next, l2);
+        }
+
+        return node;
+    }
+};

lcof/面试题26. 树的子结构/README.md

@@ -162,6 +162,53 @@ func helper(a *TreeNode, b *TreeNode) bool {
 }
 ```
 
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    bool isSubTree(TreeNode* a, TreeNode* b) {
+        if (nullptr == b) {
+            // 如果小树走到头，则表示ok了
+            return true;
+        }
+
+        if (nullptr == a) {
+            // 如果大树走到头，小树却没走到头，说明不对了
+            return false;
+        }
+
+        if (a->val != b->val) {
+            return false;
+        }
+
+        return isSubTree(a->left, b->left) && isSubTree(a->right, b->right);
+    }
+
+    bool isSubStructure(TreeNode* a, TreeNode* b) {
+        bool ret = false;
+        if (nullptr != a && nullptr != b) {
+            // 题目约定，空树不属于任何一个数的子树
+            if (a->val == b->val) {
+                // 如果值相等，才进入判定
+                ret = isSubTree(a, b);
+            }
+
+            if (false == ret) {
+                ret = isSubStructure(a->left, b);
+            }
+
+            if (false == ret) {
+                ret = isSubStructure(a->right, b);
+            }
+        }
+
+        return ret;
+    }
+};
+```
+
 ### **...**
 
 ```

lcof/面试题26. 树的子结构/Sulotion.cpp

@@ -0,0 +1,49 @@
+/**
+ * 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 isSubTree(TreeNode* a, TreeNode* b) {
+        if (nullptr == b) {
+            // 如果小树走到头，则表示ok了
+            return true;
+        }
+
+        if (nullptr == a) {
+            // 如果大树走到头，小树却没走到头，说明不对了
+            return false;
+        }
+
+        if (a->val != b->val) {
+            return false;
+        }
+
+        return isSubTree(a->left, b->left) && isSubTree(a->right, b->right);
+    }
+
+    bool isSubStructure(TreeNode* a, TreeNode* b) {
+        bool ret = false;
+        if (nullptr != a && nullptr != b) {
+            if (a->val == b->val) {
+                // 如果值相等，才进入判定
+                ret = isSubTree(a, b);
+            }
+
+            if (false == ret) {
+                ret = isSubStructure(a->left, b);
+            }
+
+            if (false == ret) {
+                ret = isSubStructure(a->right, b);
+            }
+        }
+
+        return ret;
+    }
+};