Posts tagged as “serialization”

花花酱 LeetCode 331. Verify Preorder Serialization of a Binary Tree

By zxi on April 13, 2018

Problem

题目大意：验证二叉树前序遍历的序列化是否合法。

https://leetcode.com/problems/verify-preorder-serialization-of-a-binary-tree/description/

One way to serialize a binary tree is to use pre-order traversal. When we encounter a non-null node, we record the node’s value. If it is a null node, we record using a sentinel value such as #.

     _9_
    /   \
   3     2
  / \   / \
 4   1  #  6
/ \ / \   / \
# # # #   # #

For example, the above binary tree can be serialized to the string "9,3,4,#,#,1,#,#,2,#,6,#,#", where #represents a null node.

Given a string of comma separated values, verify whether it is a correct preorder traversal serialization of a binary tree. Find an algorithm without reconstructing the tree.

Each comma separated value in the string must be either an integer or a character '#' representing nullpointer.

You may assume that the input format is always valid, for example it could never contain two consecutive commas such as "1,,3".

Example 1:
"9,3,4,#,#,1,#,#,2,#,6,#,#"
Return true

Example 2:
"1,#"
Return false

Example 3:
"9,#,#,1"
Return false

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

Solution: Recursion

If a node is not null, it must has two children, thus verify left subtree and right subtree recursively.
If a not is null, the current char must be ‘#’

Time complexity: O(n)

Space complexity: O(h)

C++

// Author: Huahua

// Running time: 5 ms

class Solution {

public:

bool isValidSerialization(string preorder) {

int pos = 0;

return isValid(preorder, pos) && pos == preorder.length();

}

private:

bool isValid(const string& s, int& pos) {

if (pos >= s.length()) return false;

if (isdigit(s[pos])) {

while (isdigit(s[pos])) ++pos;

return isValid(s, ++pos) && isValid(s, ++pos);

}

return s[pos++] == '#';

}

};

花花酱 LeetCode 652. Find Duplicate Subtrees

By zxi on December 31, 2017

652. Find Duplicate SubtreesMedium730151FavoriteShare

Given a binary tree, return all duplicate subtrees. For each kind of duplicate subtrees, you only need to return the root node of any one of them.

Two trees are duplicate if they have the same structure with same node values.

Example 1:

The following are two duplicate subtrees:

2
/
4

and

Therefore, you need to return above trees’ root in the form of a list.

Solution 1: Serialization

Time complexity: O(n^2)
Space complexity: O(n^2)

C++

// Author: Huahua

// Runtime: 29 ms

class Solution {

public:

vector<TreeNode*> findDuplicateSubtrees(TreeNode* root) {

unordered_map<string, int> counts;

vector<TreeNode*> ans;

serialize(root, counts, ans);

return ans;

}

private:

string serialize(TreeNode* root, unordered_map<string, int>& counts, vector<TreeNode*>& ans) {

if (!root) return "#";

string key = to_string(root->val) + ","

+ serialize(root->left, counts, ans) + ","

+ serialize(root->right, counts, ans);

if (++counts[key] == 2)

ans.push_back(root);

return key;

}

};

Solution 2: int id for each unique subtree

Time complexity: O(n)
Space complexity: O(n)

C++

// Author: Huahua

// Runtime: 8 ms

class Solution {

public:

vector<TreeNode*> findDuplicateSubtrees(TreeNode* root) {

unordered_map<long, pair<int,int>> counts;

vector<TreeNode*> ans;

getId(root, counts, ans);

return ans;

}

private:

int getId(TreeNode* root,

unordered_map<long, pair<int,int>>& counts,

vector<TreeNode*>& ans) {

if (!root) return 0;

long key = (static_cast<long>(static_cast<unsigned>(root->val)) << 32) +

(getId(root->left, counts, ans) << 16) +

getId(root->right, counts, ans);

auto& p = counts[key];

if (p.second++ == 0)

p.first = counts.size();

else if (p.second == 2)

ans.push_back(root);

return p.first;

}

};

花花酱 LeetCode 449. Serialize and Deserialize BST

By zxi on October 19, 2017

Problem:

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 search tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary search tree can be serialized to a string and this string can be deserialized to the original tree structure.

The encoded string should be as compact as possible.

Note: Do not use class member/global/static variables to store states. Your serialize and deserialize algorithms should be stateless.

Idea:

Binary format

serialized size: 4*n bytes, n is the number of nodes in the BST.

Time complexity: O(n)

Space complexity: O(n)

Solution:

C++

// Author: Huahua

// Runtime: 19 ~ 26 ms (<93.31%)

class Codec {

public:

// Encodes a tree to a single string.

string serialize(TreeNode* root) {

string s;

serialize(root, s);

return s;

}

// Decodes your encoded data to tree.

TreeNode* deserialize(string data) {

int pos = 0;

return deserialize(data, pos, INT_MIN, INT_MAX);

}

private:

void serialize(TreeNode* root, string& s) {

if (!root) return;

s.append(reinterpret_cast<const char*>(&root->val), sizeof(root->val));

serialize(root->left, s);

serialize(root->right, s);

}

TreeNode* deserialize(const string& s, int& pos, int curMin, int curMax) {

if (pos >= s.size()) return nullptr;

int val = *reinterpret_cast<const int*>(s.data() + pos);

if (val < curMin || val > curMax) return nullptr;

pos += sizeof(val);

TreeNode* root = new TreeNode(val);

root->left = deserialize(s, pos, curMin, val);

root->right = deserialize(s, pos, val, curMax);

return root;

}

};

Related Problems

[解题报告] LeetCode 297. Serialize and Deserialize Binary Tree

花花酱 LeetCode 297. Serialize and Deserialize Binary Tree

By zxi on September 17, 2017

Problem:

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

/ \

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.

https://leetcode.com/problems/serialize-and-deserialize-binary-tree/description/

Idea:

Recursion

Time Complexity O(n)

Solution 1: ASCII

C++

// Author: Huahua

// Running time: 39 ms

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 << "# ";

return;

}

out << root->val << " ";

serialize(root->left, out);

serialize(root->right, 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;

}

};

Solution 2: Binary

C++

// Author: Huahua

// Running time: 23 ms (beat 98.07%)

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:

enum STATUS {

ROOT_NULL = 0x0,

ROOT = 0x1,

LEFT = 0x2,

RIGHT = 0x4

};

void serialize(TreeNode* root, ostringstream& out) {

char status = 0;

if (root) status |= ROOT;

if (root && root->left) status |= LEFT;

if (root && root->right) status |= RIGHT;

out.write(&status, sizeof(char));

if (!root) return;

out.write(reinterpret_cast<char*>(&(root->val)), sizeof(root->val));

if (root->left) serialize(root->left, out);

if (root->right) serialize(root->right, out);

}

TreeNode* deserialize(istringstream& in) {

char status;

in.read(&status, sizeof(char));

if (!status & ROOT) return nullptr;

auto root = new TreeNode(0);

in.read(reinterpret_cast<char*>(&root->val), sizeof(root->val));

root->left = (status & LEFT) ? deserialize(in) : nullptr;

root->right = (status & RIGHT) ? deserialize(in) : nullptr;

return root;

}

};

// Your Codec object will be instantiated and called as such:

// Codec codec;

// codec.deserialize(codec.serialize(root));

C++ (string)

// Author: Huahua

// Running time: 23 ms (<98.13%)

class Codec {

public:

// Encodes a tree to a single string.

string serialize(TreeNode* root) {

string s;

serialize(root, s);

return s;

}

// Decodes your encoded data to tree.

TreeNode* deserialize(string data) {

int pos = 0;

return deserialize(data, pos);

}

private:

enum STATUS {

ROOT_NULL = 0x0,

ROOT = 0x1,

LEFT = 0x2,

RIGHT = 0x4

};

void serialize(TreeNode* root, string& s) {

char status = ROOT_NULL;

if (root) status |= ROOT;

if (root && root->left) status |= LEFT;

if (root && root->right) status |= RIGHT;

s.push_back(status);

if (!root) return;

s.append(reinterpret_cast<char*>(&root->val), sizeof(root->val));

if (root->left) serialize(root->left, s);

if (root->right) serialize(root->right, s);

}

TreeNode* deserialize(const string& s, int& pos) {

char status = s[pos++];

if (!status) return nullptr;

TreeNode* root = new TreeNode(0);

memcpy(&root->val, s.data() + pos, sizeof(root->val));

pos += sizeof(root->val);

root->left = (status & LEFT) ? deserialize(s, pos) : nullptr;

root->right = (status & RIGHT) ? deserialize(s, pos) : nullptr;

return root;

}

};