题解 | #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:

    void SerializeFun(TreeNode* root, string& str) {
        if (root == nullptr) {
            str += '#';
            return;
        }
        string temp = to_string(root->val);
        str += temp + ',';
        SerializeFun(root->left, str);
        SerializeFun(root->right, str);
    }
    char* Serialize(TreeNode* root) {
        if (root == nullptr)
            return "#";
        string res;
        SerializeFun(root, res);
        char* charRes = new char[res.length() + 1];
        strcpy(charRes, res.c_str());
        charRes[res.length()] = '\0';
        return charRes;
    }
    TreeNode* DeserializeFun(char** str) {
        if (**str == '#') {
            (*str)++;
            return nullptr;
        }
        int val = 0;
        while (**str != ',' &&** str != '\0') {
            val = val * 10 + ((**str) - '0');
            (*str)++;
        }
        TreeNode* root = new TreeNode(val);
        if (**str == '\0')
            return root;
        else
            (*str)++;
        root->left = DeserializeFun(str);
        root->right = DeserializeFun(str);
        return root;
    }
    TreeNode* Deserialize(char* str) {
        if(str == "#")
            return nullptr;
        return DeserializeFun(&str);
    }
};