297. Serialize and Deserialize Binary Tree¶
Tags: Hard
Tree
Links: https://leetcode.com/problems/serialize-and-deserialize-binary-tree/
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.
Example:
You may serialize the following tree:
1
/ \
2 3
/ \
4 5
as "[1,2,3,null,null,4,5]"
Clarification: The above format is the same as how LeetCode 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.
在《树——二叉树的构建》里面进行了总结。这道题其实就是一本通-1340:【例3-5】扩展二叉树的一个翻版,只不过现在需要自己手动实现扩展二叉树的输出,并且有一个很重要的不同点,相比于一本通1340里的单个字母,二叉树里的节点存储的数值可能不止一位,所以需要用一个特殊的标记符号#
来对数据进行分隔,用.
代表空节点。
序列化部分采用前序遍历的递归实现,注意需要去掉末尾的#
符号,用pos
记录处理到序列的哪个位置,每次指向第一个数字,然后递归构建左右子树。
/**
* 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:
// Encodes a tree to a single string.
string serialize(TreeNode* root) {
std::ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
string res;
if (!root) return res;
preorderTraversal(root, res);
return res.substr(0, res.size() - 1);
}
void preorderTraversal(TreeNode *root, string & res)
{
if (!root) {
res += ".#"; return;
}
res += to_string(root -> val);
res.push_back('#');
preorderTraversal(root -> left, res);
preorderTraversal(root -> right, res);
}
// Decodes your encoded data to tree.
TreeNode* deserialize(string data) {
if (data.empty()) return NULL;
int pos = 0;
TreeNode *root;
build(data, root, pos);
return root;
}
void build(string & data, TreeNode *&root, int &pos)
{
int nextPos = data.find("#", pos);
if (nextPos == string::npos) {
string tmp = data.substr(pos);
if (tmp.size() == 1 && tmp[0] == '.') root = NULL;
else root = new TreeNode(stoi(tmp));
return;
}
string tmp = data.substr(pos, nextPos - pos);
pos = nextPos + 1;
if (tmp[0] == '.') root = NULL;
else {
root = new TreeNode(stoi(tmp));
build(data, root -> left, pos);
build(data, root -> right, pos);
}
}
};
// Your Codec object will be instantiated and called as such:
// Codec codec;
// codec.deserialize(codec.serialize(root));