剑指OFFER 序列化二叉树

弄了半天在剑指OFFER OJ上无法通过(猜测可能是因为剑指OFFER上使用的是char类型的指针,导致有一些编译的不一致问题),同样的代码在leetcode上通过了

代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Codec {
public:
    int pos = 0;
    string str;
    string res;

    void recur_from_order(TreeNode* &node)
    {
        if(str[pos] == '#'){
            node = NULL;
            return;
        }else{
            string str_num;
            while(str[pos] != '!')
            {
                str_num += str[pos];
                pos++;
            }
            //int num = stoi(str_num);
            int num = atoi(str_num.c_str());
            node = (TreeNode*)malloc(sizeof(TreeNode));
            node->val = num;
        }
        pos++;
        recur_from_order(node->left);
        pos++;
        recur_from_order(node->right);
    }
    
    void recur_to_order(TreeNode *node)
    {
        if(node == NULL){
            res += '#';
            return;
        }else{
            res += to_string(node->val);
            res += '!';
        }
        recur_to_order(node->left);
        recur_to_order(node->right);
    }
    
    // Encodes a tree to a single string.
    string serialize(TreeNode* root) {
        recur_to_order(root);
        return res;
    }

    // Decodes your encoded data to tree.
    TreeNode* deserialize(string data) {
        str = data;
        TreeNode* ret;
        recur_from_order(ret);
        return ret;
    }
};

// Your Codec object will be instantiated and called as such:
// Codec codec;
// codec.deserialize(codec.serialize(root));

采用的先序遍历的方式,先写序列化的代码,就是先序遍历一遍,遇到结点就记录到res中.

反序列化稍微复杂一点,聚焦于两个状态,一个是结点,一个是字符串的位置指针,通过这两个状态来构建二叉树.