add 297

README.md

@@ -195,6 +195,7 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
 | 289 |[Game of Life](https://leetcode.com/problems/game-of-life/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/289.%20Game%20of%20Life.md)|Medium|  |
 | 290 |[Word Pattern](https://leetcode.com/problems/word-pattern)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/290.%20Word%20Pattern.md)|Easy|  |
 | 292 |[Nim Game](https://leetcode.com/problems/nim-game)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/292.%20Nim%20Game.md)|Easy|  |
+| 297 |[Serialize and Deserialize Binary Tree](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/)|[C++](solutions/297.%20Serialize%20and%20Deserialize%20Binary%20Tree.md)|Hard|  |
 | 300 |[Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/300.%20Longest%20Increasing%20Subsequence.md)|Medium|  |
 | 303 |[Range Sum Query - Immutable](https://leetcode.com/problems/range-sum-query-immutable)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/303.%20Range%20Sum%20Query%20-%20Immutable.md)|Easy|  |
 | 304 |[Range Sum Query 2D - Immutable](https://leetcode.com/problems/range-sum-query-2d-immutable/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/304.%20Range%20Sum%20Query%202D%20-%20Immutable.md)|Medium|  |

solutions/297. Serialize and Deserialize Binary Tree.md

@@ -0,0 +1,135 @@
+# [297. Serialize and Deserialize Binary Tree](https://leetcode.com/problems/serialize-and-deserialize-binary-tree/)
+
+# 思路
+对二叉树进行序列化和去序列化的操作。
+
+## 思路一
+本质上就是遍历二叉树，那我们考虑用递归前序遍历二叉树。
+* 序列化：
+    * 如果节点为空则添加"# "（用空格作为分隔符，这样就可以用字符串流）；
+    * 否则添加节点值（要紧跟一个空格）再递归进入左右孩子。
+
+* 去序列化：
+    * 如果当前字符为"#"，则返回空；
+    * 否则根据字符串的值新建节点，然后递归求其左右孩子。
+
+我们可以用输入/输出字符串流`istringstream`/`ostringstream`来进行处理，详见代码。
+
+## 思路二
+我们还可以用非递归的层序遍历，这样就类似LeetCode使用的方法了。
+
+* 序列化：
+    * 将根节点入队列，循环直到队列为空：出队首元素，若其为空则添加"# "；否则添加节点值（要紧跟一个空格），另外还要将其左右孩子入队（即使孩子为空）。
+
+* 去序列化：
+    * 先处理得到根节点，然后将根节点入队，循环直到队列空：处理得到队首节点的左孩子，即将队首元素的左指针指向当前处理得到的节点，若这个左孩子不为空还要将其入队；再用类似的方法处理得到队首节点的右孩子；最后将队首元素出队。
+
+同样用输入/输出字符串流`istringstream`/`ostringstream`来进行处理，详见代码。
+
+本质都是遍历二叉树，所以两个思路时空复杂度均为O(n)
+
+# C++
+## 思路一
+``` C++
+class Codec {
+private:
+    void serialize(TreeNode* root, ostringstream &out){
+        if(!root) out << "# ";
+        else{
+            out << root -> val << " ";
+            serialize(root -> left, out);
+            serialize(root -> right, out);
+        }
+    }
+    TreeNode* deserialize(istringstream &in){
+        string val;
+        in >> val;
+        if(val == "#") return NULL;
+        
+        TreeNode *root = new TreeNode(stoi(val));
+        root -> left = deserialize(in);
+        root -> right = deserialize(in);
+        return root;
+    }
+        
+public:
+
+    // Encodes a tree to a single string.
+    string serialize(TreeNode* root) {
+        if(!root) return "";
+        ostringstream out;
+        serialize(root, out);
+        return out.str();   
+    }
+
+    // Decodes your encoded data to tree.
+    TreeNode* deserialize(string data) {
+        if(!data.size()) return NULL;
+        
+        istringstream in(data);
+        return deserialize(in);
+    }
+};
+```
+
+## 思路二
+``` C++
+class Codec {
+public:
+    // Encodes a tree to a single string.
+    string serialize(TreeNode* root) {
+        if(!root) return "";
+        ostringstream out;
+        
+        TreeNode *p;
+        queue<TreeNode *>q;
+        q.push(root);
+        
+        while(!q.empty()){
+            p = q.front(); q.pop();
+            if(!p) out << "# ";
+            else{
+                out << p -> val << " ";
+                q.push(p -> left);
+                q.push(p -> right);
+            }
+        }
+        return out.str();
+    }
+
+    // Decodes your encoded data to tree.
+    TreeNode* deserialize(string data) {
+        int len = data.size();
+        if(!len) return NULL;
+        
+        queue<TreeNode *>q;
+        TreeNode *p = NULL, *root = NULL;
+        
+        istringstream in(data);
+        string val;
+        
+        // 根节点
+        in >> val;
+        p = new TreeNode(stoi(val)); root = p;
+        q.push(p);
+        
+        while(!q.empty()){
+            in >> val; // 左孩子
+            if(val != "#"){
+                p = new TreeNode(stoi(val)); 
+                q.push(p);
+                q.front() -> left = p;
+            }
+            
+            in >> val; // 右孩子
+            if(val != "#"){
+                p = new TreeNode(stoi(val)); 
+                q.push(p);
+                q.front() -> right = p;
+            }
+            q.pop();
+        }
+        return root;
+    }
+};
+```