题目描述
Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.
Design an algorithm to serialize and deserialize a binary tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary tree can be serialized to a string and this string can be deserialized to the original tree structure.
题目大意
实现将一颗二叉树序列化和逆序列化的操作(序列化的结果不定,可以自己决定,但要求序列化之后再逆序列化之后的二叉树要与之前的一样)。
示例
E1
解题思路
序列化:dfs遍历二叉树,将二叉树中的结果转为字符串,树结点之间用‘,’分隔开,遇到NULL结点用‘#’表示。
逆序列化:依次遍历字符串,同时dfs建立二叉树。
复杂度分析
时间复杂度:O(N)
空间复杂度:O(N)
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: // 将二叉树序列化 string serialize(TreeNode* root) { if(root == NULL) return "#"; return to_string(root->val) + ',' + serialize(root->left) + ',' + serialize(root->right); } // 将字符串逆序列化 TreeNode* deserialize(string data) { return solve(data); } TreeNode* solve(string& data) { if(data[0] == '#') { if(data.length() > 1) data = data.substr(2); return NULL; } else { TreeNode* root = new TreeNode(helper(data)); root->left = solve(data); root->right = solve(data); return root; } } private: // 查找当前结点所对应的整数并将其表示出来,同时使字符串更新 int helper(string& s) { int pos = s.find(','); int res = stoi(s.substr(0, pos)); s = s.substr(pos + 1); return res; } }; // Your Codec object will be instantiated and called as such: // Codec codec; // codec.deserialize(codec.serialize(root));