61、剑指offer--序列化二叉树

题目描述

请实现两个函数，分别用来序列化和反序列化二叉树

 

解题思路：序列化的时候，遇到无子结点，则填#来占位；对于反序列化时，每次遇到#返回空

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};
*/
class Solution {
public:
    void Serialize(TreeNode *root, string &str)
    {
        if(root == NULL)
        {
            str += '#';//用#代表null
            return;//返回上一层
        }
        else
        {
            string r = to_string(root->val);//值转字符串
           str += r;//字符串拼接
           str += ',';//数字后用，进行节点分隔
           Serialize(root->left,str);
           Serialize(root->right,str);
        }
    }
    char* Serialize(TreeNode *root) {
        if(root == NULL)
            return NULL;
        string str;
        Serialize(root, str);
        //字符串转char *
        char *ret = new char[str.length()+1];
        for(int i=0;i<str.length();i++)
        {
            ret[i] = str[i];
        }
        ret[str.length()] = '\0';
        return ret;
    }
    TreeNode *deserialize(char **str)
    {
        if(**str == '#')//空的时候返回
        {
            ++(*str);
            return NULL;
        }
        int num = 0;
        while(**str != '\0' && **str != ',')
        {
            num = num*10 + ((**str) - '0'); //求出每个节点的值
            ++(*str);
        }
        TreeNode *root = new TreeNode(num);
        if(**str == '\0')
            return root;
        else
            (*str)++;//跨过','
        root->left = deserialize(str);
        root->right = deserialize(str);
        return root;
    }
    TreeNode* Deserialize(char *str) {
        if(str == NULL)
            return NULL;
        TreeNode *res = deserialize(&str);
        return res;
    }
};