更新了55-1，55-2和63题目的Cpp解法 (#338)

* 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

* 更新了[55-1 二叉树的深度][55-2 平衡二叉树][63 股票的最大利润]的cpp解法.以上解法均在leetcode上验证通过,基本双优

* 更新了[25.合并两个排序链表]的信息

Co-authored-by: junfeng.fj <junfeng.fj@alibaba-inc.com>
Co-authored-by: Yang Libin <szuyanglb@outlook.com>

Author: 3 people

lcof/面试题55 - I. 二叉树的深度/README.md

@@ -94,6 +94,23 @@ var maxDepth = function (root) {
 };
 ```
 
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int maxDepth(TreeNode* root) {
+        if (nullptr == root) {
+            return 0;
+        }
+
+        int left = maxDepth(root->left);
+        int right = maxDepth(root->right);
+        return std::max(left, right) + 1;
+    }
+};
+```
+
 ### **...**
 
 ```

lcof/面试题55 - I. 二叉树的深度/Solution.cpp

@@ -0,0 +1,22 @@
+/**
+ * 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:
+    int maxDepth(TreeNode* root) {
+        if (nullptr == root) {
+            return 0;
+        }
+
+        int left = maxDepth(root->left);
+        int right = maxDepth(root->right);
+        return std::max(left, right) + 1;
+    }
+};

lcof/面试题55 - II. 平衡二叉树/README.md

@@ -120,6 +120,46 @@ function getDepth(node) {
 }
 ```
 
+### **C++**
+
+```cpp
+class Solution {
+public:
+    bool isBalanced(TreeNode* root, int* depth) {
+        if (root == nullptr) {
+            *depth = 0;
+            return true;
+        }
+
+        int left = 0;
+        int right = 0;
+        if (isBalanced(root->left, &left) && isBalanced(root->right, &right)) {
+            // 这样做的优势是，不用每次都计算单个子树的深度
+            int diff = left - right;
+            if (diff > 1 || diff < -1) {
+                // 如果有一处不符合 -1 < diff < 1，则直接返回false
+                return false;
+            } else {
+                // 如果符合，则记录当前深度，然后返回上一层继续计算
+                *depth = max(left, right) + 1;
+                return true;
+            }
+        }
+
+        return false;    // 如果
+    }
+
+    bool isBalanced(TreeNode* root) {
+        if (!root) {
+            return true;
+        }
+
+        int depth = 0;
+        return isBalanced(root, &depth);
+    }
+};
+```
+
 ### **...**
 
 ```

lcof/面试题55 - II. 平衡二叉树/Solution.cpp

@@ -0,0 +1,45 @@
+/**
+ * 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 isBalanced(TreeNode* root, int* depth) {
+        if (root == nullptr) {
+            *depth = 0;
+            return true;
+        }
+
+        int left = 0; 
+        int right = 0;
+        if (isBalanced(root->left, &left) 
+            && isBalanced(root->right, &right)) {
+            int diff = left - right;
+            if (diff > 1 || diff < -1) {
+                // 如果有一处不符合 -1 < diff < 1，则直接返回false
+                return false;
+            } else {
+                // 如果符合，则记录当前深度，然后返回上一层继续计算
+                *depth = max(left, right) + 1;
+                return true;
+            }
+        }
+
+        return false;    // 如果
+    }
+
+    bool isBalanced(TreeNode* root) {
+        if (!root) {
+            return true;
+        }
+
+        int depth = 0;
+        return isBalanced(root, &depth);
+    }
+};

lcof/面试题63. 股票的最大利润/README.md

@@ -82,6 +82,37 @@ var maxProfit = function (prices) {
 };
 ```
 
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int maxProfit(vector<int>& prices) {
+        if (prices.size() < 2) {
+            return 0;    // 如果小于两个，直接返回0值
+        }
+
+        int curMin = prices[0];
+        int maxDiff = prices[1] - prices[0];
+
+        // 贪心循环，记录当前最小值，和最大diff值
+        for (int i = 2; i < prices.size(); i++) {
+            if (curMin > prices[i-1]) {
+                curMin = prices[i-1];
+            }
+
+            int diff = prices[i] - curMin;
+            if (maxDiff < diff) {
+                maxDiff = diff;
+            }
+        }
+
+        // 根据题意，如果是负数的话，则返回0值
+        return maxDiff > 0 ? maxDiff : 0;
+    }
+};
+```
+
 ### **...**
 
 ```

lcof/面试题63. 股票的最大利润/Solution.cpp

@@ -0,0 +1,24 @@
+class Solution {
+public:
+    int maxProfit(vector<int>& prices) {
+        if (prices.size() < 2) {
+            return 0;
+        }
+
+        int curMin = prices[0];
+        int maxDiff = prices[1] - prices[0];
+
+        for (int i = 2; i < prices.size(); i++) {
+            if (curMin > prices[i-1]) {
+                curMin = prices[i-1];
+            }
+
+            int diff = prices[i] - curMin;
+            if (maxDiff < diff) {
+                maxDiff = diff;
+            }
+        }
+
+        return maxDiff > 0 ? maxDiff : 0;
+    }
+};