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:
    string s;
    // 用前序遍历
    void ser(TreeNode* root, string& s) {
        if(root==nullptr) {
            s+='#';
            s+=',';
            return;
        }

        s+=to_string(root->val);
        s+=',';
        ser(root->left,s);
        ser(root->right,s);
    }

    char* Serialize(TreeNode *root) {    
        ser(root,s);
        return const_cast<char*>(s.c_str());
    }

    TreeNode* Deserialize(char *str) {
        string data(str);
//         cout<<"s: "<<data<<endl;
        int index = 0;
        return des(data,index);
    }

    TreeNode* des(string& s, int& index) {
        if(index==s.size() || index>s.size()) {
            return nullptr;
        }
        if(s[index]=='#') {
            index++;
            index++;
            return nullptr;
        }
        int sum = 0;
        while(s[index]!=',') {
            sum=sum*10+s[index]-'0';
            index++;
        }
        index++;
        TreeNode* root = new TreeNode(sum);
        root->left=des(s,index);
        root->right=des(s,index);
        return root;
    }
};