Serialize and Deserialize N-ary Tree in C++

C++Server Side ProgrammingProgramming

Suppose we have one N-ary tree and we have to serialize and deserialize them. As we know that the serialization is the process of converting a data structure or object into a sequence of bits so we can store them in a file or memory buffer, and that can be reconstructed later in the same or another computer environment.

Here we have to devise an algorithm to serialize and deserialize an N-ary tree. The N-ary tree is a rooted tree in which each node has no more than N children.

So, if the input is like

then the output will be Serialize: 1 #3 2 4 #5 6 ##### and Deserialized Tree: 1[3[5, 6, ], 2, 4, ]

To solve this, we will follow these steps −

  • Define a function createVector(), this will take s,

  • Define one 2D array ret

  • Define an array tempv

  • temp := empty string

  • for initialize i := 0, when i < size of s, update (increase i by 1), do −

    • if s[i] is not equal to blank space and s[i] is not equal to '#', then −

      • temp := temp + s[i]

    • otherwise when s[i] is same as blank string, then −

      • insert temp s integer at the end of tempv

      • temp := empty string

    • otherwise when s[i] is same as '#', then −

      • insert tempv at the end of ret

      • temp := empty string

      • clear tempv

  • while (not ret is empty and last element of ret is same as 0), do −

    • delete last element from ret

  • return ret

  • Define a function serialize(), this will take root,

  • ret := empty string

  • if not root is non-zero, then −

    • return ret

  • Define one queue q

  • insert root into q

  • rret := ret concatenate val of root

  • ret := ret concatenate space

  • ret := ret concatenate "#"

  • while (not q is empty), do −

    • curr = first element of q

    • delete element from q

    • for initialize i := 0, when i < size of the children array of curr, update (increase i by 1), do −

      • if children[i] of curr, then −

        • ret := ret concatenate children[i] of curr

        • insert children[i] of curr into q

      • ret := ret + blank string

    • ret := ret concatenate "#"

  • return ret

  • Define a function deserialize(), this will take data,

  • if size of data is same as 0, then −

    • return null

  • Define one 2D array v := createVector(data)

  • ret := new node with value v[0, 0]

  • Define one queue q

  • insert ret into q

  • i := 1

  • while (not q is empty and i − size of v), do −

    • curr = first element of q

    • delete element from q

    • for initialize j := 0, when j − size of v[i],update (increase j by 1), do −

      • node := v[i, j]

      • temp = new node with value node

      • insert temp at the end of children of curr

      • insert temp into q

    • (increase i by 1)

  • return ret

Example 

Let us see the following implementation to get better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Node {
public:
   int val;
   vector<Node*> children;
   Node() {}
   Node(int _val) {
      val = _val;
   }
   Node(int _val, vector<Node*> _children) {
      val = _val;
      children = _children;
   }
};
string n_ary_to_str(Node *root){
   string ret = "";
   if(root){
      ret = ret + to_string(root->val);
      if(root->children.size() > 0){
         ret += "[";
         for(Node* child : root->children){
            ret += n_ary_to_str(child) + ", ";
         }
         ret += "]";
      }
   }
   return ret;
}
class Codec {
public:
   vector<vector<int>>createVector(string s) {
      vector<vector<int>> ret;
      vector<int> tempv;
      string temp = "";
      for (int i = 0; i < s.size(); i++) {
         if (s[i] != ' ' && s[i] != '#') {
            temp += s[i];
         }
         else if (s[i] == ' ') {
            tempv.push_back(stoi(temp));
            temp = "";
         }
         else if (s[i] == '#') {
            ret.push_back(tempv);
            temp = "";
            tempv.clear();
         }
      }
      while (!ret.empty() && ret.back().size() == 0)
      ret.pop_back();
      return ret;
   }
   string serialize(Node *root) {
      string ret = "";
      if (!root)
         return ret;
      queue<Node *> q;
      q.push(root);
      ret += to_string(root->val);
      ret += " ";
      ret += "#";
      while (!q.empty()) {
         Node *curr = q.front();
         q.pop();
         for (int i = 0; i < curr->children.size(); i++) {
            if (curr->children[i]) {
               ret += to_string(curr->children[i]->val);
               q.push(curr->children[i]);
            }
            ret += " ";
         }
         ret += "#";
      }
      return ret;
   }
   Node *deserialize(string data) {
      Node *ret;
      if (data.size() == 0)
         return NULL;
         vector<vector<int>> v = createVector(data);
         ret = new Node(v[0][0]);
         queue<Node *> q;
         q.push(ret);
         int i = 1;
         while (!q.empty() && i < v.size()) {
            Node *curr = q.front();
            q.pop();
            for (int j = 0; j < v[i].size(); j++) {
               int node = v[i][j];
                  Node *temp = new Node(node);
                  curr->children.push_back(temp);
                  q.push(temp);
               }
               i++;
            }
            return ret;
         }
 };
main() {
   Codec ob;
   Node n5(5), n6(6);
   Node n3(3); n3.children.push_back(&n5); n3.children.push_back(&n6);
   Node n2(2), n4(4);
   Node n1(1); n1.children.push_back(&n3); n1.children.push_back(&n2);
   n1.children.push_back(&n4);
   cout << "Given Tree: " << n_ary_to_str(&n1) << endl;
   string ser = ob.serialize(&n1);
   cout << "Serialize: " << ser << endl;
   Node *deser = ob.deserialize(ser);
   cout << "Deserialized Tree: " << n_ary_to_str(deser);
}

Input

Node n5(5), n6(6);
Node n3(3); n3.children.push_back(&n5); n3.children.push_back(&n6);
Node n2(2), n4(4);
Node n1(1); n1.children.push_back(&n3); n1.children.push_back(&n2);
n1.children.push_back(&n4);

Output

Given Tree: 1[3[5, 6, ], 2, 4, ]
Serialize: 1 #3 2 4 #5 6 #####
Deserialized Tree: 1[3[5, 6, ], 2, 4, ]
raja
Published on 21-Jul-2020 08:53:21
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