面试题37: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

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};
*/
class Solution {
public:
    char* Serialize(TreeNode *root) {
        if(root == nullptr)
            return nullptr;
        string str;
        helpS(root,str);
        char* res = new char[str.length()+1];
        int i = 0;
        for(; i < str.length(); ++i)
            res[i] = str[i];
        res[i] = '\0';
        return res;
    }

    void helpS(TreeNode* root, string& str)
    {
        if(root == nullptr)
        {
            str +='#';
            return;
        }
        string ch = to_string(root->val);
        str += ch;
        str += ',';
        helpS(root->left,str);
        helpS(root->right,str);
    }

    TreeNode* Deserialize(char *str) {
        if(str == nullptr)
            return nullptr;
        TreeNode* res = helpD(&str);
        return res;
    }

    TreeNode* helpD(char** str)//*str是指向字符串中str这个字符的指针,因为指针要往后移,要指向下一个字符,也就是说过程中要改变指针,所以应该传这个指针的地址,所以是二级指针
    {
        if(**str == '#')
        {
            ++ (*str);
            return nullptr;
        }
        int num = 0;
        while(**str != '\0' && **str != ',')
        {
            num = num*10 + ((**str) - '0');
            ++(*str);
        }
        TreeNode* root = new TreeNode(num);//建立根节点
        if((**str) == '\0')
            return root;
        else
            (*str) ++;
        root->left = helpD(str);
        root->right = helpD(str);
        return root;
    }
};

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转载自blog.csdn.net/chineseqsc/article/details/81603900