[LeetCode] Serialize and Deserialize Binary Tree 二叉樹的序列化和去序列化
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.
For example, you may serialize the following tree
1 / \ 2 3 / \ 4 5
as "[1,2,3,null,null,4,5]"
, just the same as how LeetCode OJ serializes a binary tree. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.
Note: Do not use class member/global/static variables to store states. Your serialize and deserialize algorithms should be stateless.
Credits:
Special thanks to @Louis1992 for adding this problem and creating all test cases.
這道題讓我們對二叉樹進行序列化和去序列化的操作。序列化就是將一個數據結構或物體轉化為一個位序列,可以存進一個檔案或者記憶體緩衝器中,然後通過網路連線在相同的或者另一個電腦環境中被還原,還原的過程叫做去序列化。現在讓我們來序列化和去序列化一個二叉樹,並給了我們例子。這題有兩種解法,分別為先序遍歷的遞迴解法和層序遍歷的非遞迴解法。先來看先序遍歷的遞迴解法,非常的簡單易懂,我們需要接入輸入和輸出字串流istringstream和ostringstream,對於序列化,我們從根節點開始,如果節點存在,則將值存入輸出字串流,然後分別對其左右子節點遞迴呼叫序列化函式即可。對於去序列化,我們先讀入第一個字元,以此生成一個根節點,然後再對根節點的左右子節點遞迴呼叫去序列化函式即可,參見程式碼如下:
解法一:
class Codec { public: // Encodes a tree to a single string. string serialize(TreeNode* root) { ostringstream out; serialize(root, out); return out.str(); } // Decodes your encoded data to tree. TreeNode* deserialize(string data) { istringstream in(data); return deserialize(in); } private: void serialize(TreeNode *root, ostringstream &out) { if (root) { out << root->val << ' '; serialize(root->left, out); serialize(root->right, out); } else { out << "# "; } } TreeNode* deserialize(istringstream &in) { string val; in >> val; if (val == "#") return nullptr; TreeNode *root = new TreeNode(stoi(val)); root->left = deserialize(in); root->right = deserialize(in); return root; } };
另一種方法是層序遍歷的非遞迴解法,這種方法略微複雜一些,我們需要藉助queue來做,本質是BFS演算法,也不是很難理解,就是BFS演算法的常規套路稍作修改即可,參見程式碼如下:
解法二:
class Codec { public: // Encodes a tree to a single string. string serialize(TreeNode* root) { ostringstream out; queue<TreeNode*> q; if (root) q.push(root); while (!q.empty()) { TreeNode *t = q.front(); q.pop(); if (t) { out << t->val << ' '; q.push(t->left); q.push(t->right); } else { out << "# "; } } return out.str(); } // Decodes your encoded data to tree. TreeNode* deserialize(string data) { if (data.empty()) return nullptr; istringstream in(data); queue<TreeNode*> q; string val; in >> val; TreeNode *res = new TreeNode(stoi(val)), *cur = res; q.push(cur); while (!q.empty()) { TreeNode *t = q.front(); q.pop(); if (!(in >> val)) break; if (val != "#") { cur = new TreeNode(stoi(val)); q.push(cur); t->left = cur; } if (!(in >> val)) break; if (val != "#") { cur = new TreeNode(stoi(val)); q.push(cur); t->right = cur; } } return res; } };
參考資料: