day18--序列化二叉树

遍历二叉树，记录每个节点，再以同样的方式遍历就可以还原二叉树

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};
*/
#include <string>
class Solution {
public:
    void Serializef(TreeNode *root, string& str){
        if(!root){ str+="#"; return ; }
        string tmp=to_string(root->val);
        str += tmp + "!";//！区分节点
        Serializef(root->left, str); //左子树
        Serializef(root->right, str); //右子树
    }
    char* Serialize(TreeNode *root) {    
        if(!root) return "#";//#表示空节点
        string res;
        Serializef(root, res);
        //str转化为char
        char* charres=new char[res.length()+1];
        strcpy(charres, res.c_str());
        charres[res.length()]='\0';
        return charres;
    }
    TreeNode* Deserializef(char** str){//**:表取值
        if(**str == '#'){ (*str)++; return NULL; }
        int val=0;
        while(**str != '!' && **str != '\0'){//转换为数字
            val = val * 10 + ((**str) - '0');
            (*str)++;
        }
        TreeNode* root=new TreeNode(val);
        if(** str == '\0') return root;//序列化到底了构建完成
        else (*str)++;
        //前序遍历
        root->left=Deserializef(str);
        root->right=Deserializef(str);
        return root;
    }
    TreeNode* Deserialize(char *str) {
        if(str == "#")return NULL;
        TreeNode* res=Deserializef(&str);
        return res;
    }
};