序列化

思路：

1、d对于经常调用string 与 int之间转换的代码 一定要建立函数 美观 好使；
2、 先序遍历，保存格式为： 数字+“ ” 空节点为 “# ”
3、 将string处理为queue 方便处理节点后直接弹出
4、遍历queue中的节点，遍历到时弹出，为“#”返回空，否则按先序遍历的方式递归遍历

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */


class Solution {
public:
    /**
     * This method will be invoked first, you should design your own algorithm 
     * to serialize a binary tree which denote by a root node to a string which
     * can be easily deserialized by your own "deserialize" method later.
     */
    int strtoint(string &svalue){
        stringstream ss;
        ss << svalue;
        int ivalue;
        ss >> ivalue;
        return ivalue;
    }
    
    string inttostr(int &ivalue){
        stringstream ss;
        ss << ivalue;
        string svalue;
        ss >> svalue;
        return svalue;
    }
    
    string serialize(TreeNode * root) {
        if(root == NULL)    //基本情况 而且处理了头结点为空的问题 很6
            return "# ";
        
        string svalue = inttostr(root->val);
        string res = svalue+" ";
        
        res += serialize(root->left);
        res += serialize(root->right);
        return res;
    }

    TreeNode * deserialize(string &data) {
        // write your code here
        queue<string> p;
        stringstream input(data);
        string temp;
        while(input >> temp){
            p.push(temp);
            temp.clear();
        }
        return deserializehelp(p);
    }
    TreeNode* deserializehelp(queue<string> &p){
        string svalue = p.front();
        p.pop();
        if(svalue == "#"){
            return NULL;
        }
        int ivalue = strtoint(svalue);
        
        TreeNode* root = new TreeNode(ivalue);
        root->left = deserializehelp(p);
        root->right = deserializehelp(p);
        return root;
    }
};

刷题：https://www.lintcode.com/problem/serialize-and-deserialize-binary-tree/