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.
这道题让我们对二叉树进行序列化和去序列化的操作。序列化就是将一个数据结构或物体转化为一个位序列，可以存进一个文件或者内存缓冲器中，然后通过网络连接在相同的或者另一个电脑环境中被还原，还原的过程叫做去序列化。现在让我们来序列化和去序列化一个二叉树，并给了我们例子。这题有两种解法，分别为先序遍历的递归解法和层序遍历的非递归解法。先来看先序遍历的递归解法，非常的简单易懂，我们需要接入输入和输出字符串流istringstream和ostringstream，对于序列化，我们从根节点开始，如果节点存在，则将值存入输出字符串流，然后分别对其左右子节点递归调用序列化函数即可。对于去序列化，我们先读入第一个字符，以此生成一个根节点，然后再对根节点的左右子节点递归调用去序列化函数即可，参见代码如下：
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;
}
};
另一种方法是层序遍历的非递归解法，这种方法略微复杂一些，我们需要借助queue来做，本质是BFS算法，也不是很难理解，就是BFS算法的常规套路稍作修改即可，参见代码如下：
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;
}
};
参考资料：
https://leetcode.com/problems/serialize-and-deserialize-binary-tree/
LeetCode All in One 题目讲解汇总(持续更新中…)
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