Binary Tree Serialization.cpp

#include<iostream>
#include<unordered_map>
#include<vector>
#include<queue>
#include<map>
#include<algorithm>
#include<cmath>
#include<set>
#include<stack>
#include<sstream>

using namespace std;

class TreeNode {
public:
 int val;
 TreeNode *left, *right;
 TreeNode(int val) {
 this->val = val;
 this->left = this->right = NULL;
 }
};


class Solution {
public:
 /**
 * This method will be invoked first, you should design your own algorithm
 * to serialize a binary tree which denote by a root node to a string which
 * can be easily deserialized by your own "deserialize" method later.
 */
 
 void help_serialize(TreeNode *p, string& result)
 {
 if (!p) {
 result +="# ";
 } else {
 result = result + to_string(p->val) + " ";
 help_serialize(p->left, result);
 help_serialize(p->right, result);
 }
 }
 
 string serialize(TreeNode *root) {
 // write your code here
 string result("");
 help_serialize(root, result);
 return result;
 }
 
 bool readNextToken(int& token, stringstream &ss, bool &isNumber)
 {
 string s;
 ss >> s;
 if(s.length() == 0) return false;
 if(s == "#")
 {
 isNumber = false;
 return true;
 }
 token = stoi(s, nullptr);
 isNumber = true;
 return true;
 }
 
 TreeNode* readBSTHelper(stringstream &ss) {
 int token;
 bool isNumber;
 TreeNode *p(nullptr);
 if (!readNextToken(token, ss, isNumber))
 return nullptr;
 if (isNumber) {
 p = new TreeNode(token);
 p->left = readBSTHelper(ss);
 p->right = readBSTHelper(ss);
 }
 return p;
 }
 
 /**
 * This method will be invoked second, the argument data is what exactly
 * you serialized at method "serialize", that means the data is not given by
 * system, it's given by your own serialize method. So the format of data is
 * designed by yourself, and deserialize it here as you serialize it in
 * "serialize" method.
 */
 TreeNode *deserialize(string data) {
 // write your code here
 stringstream ss(data);
 TreeNode* root = readBSTHelper(ss);
 return root;
 }
 
};

int main()
{
 TreeNode *n1 = new TreeNode(1);
 TreeNode *n2 = new TreeNode(2);
 TreeNode *n3 = new TreeNode(20);
 TreeNode *n4 = new TreeNode(15);
 TreeNode *n5 = new TreeNode(7);
 //n1 -> left = n2;
 n1 ->right = n2;
 //n3 ->left = n4;
 //n3 ->right = n5;
 
 Solution s;
 string res = s.serialize(n1);
 cout << res << endl;
 TreeNode* root = s.deserialize(res);
 cout << s.serialize(root);
 return 0;
}