Encode N-ary Tree to Binary Tree in C++

Suppose we have a N-ary tree. We have to encode that tree into one binary. We also have to make deserializer to deserialize the binary tree to N-ary tree.
So, if the input is like
then the output will be
To solve this, we will follow these steps −
Define a function encode(), this will take root,
Define a function encode(), this will take root,
if root is valid, then −return null
if root is valid, then −
return null
return null
node = new tree node with value of root
node = new tree node with value of root
if size of children of root is not 0, then −left of node := encode(children[0] of root)
if size of children of root is not 0, then −
left of node := encode(children[0] of root)
left of node := encode(children[0] of root)
curr = left of node
curr = left of node
for initialize i := 1, when i < size of children of root, update (increase i by 1), do −right of node := encode(children[i] of root)curr := right of curr
for initialize i := 1, when i < size of children of root, update (increase i by 1), do −
right of node := encode(children[i] of root)
right of node := encode(children[i] of root)
curr := right of curr
curr := right of curr
return node
return node
Define a function decode(), this will take root,
Define a function decode(), this will take root,
if root is not present, then −return NULL
if root is not present, then −
return NULL
return NULL
node := new node with val of root
node := new node with val of root
curr := left of root
curr := left of root
while curr is non-zero, do −insert decode(curr) at the end of children of nodecurr := right of curr
while curr is non-zero, do −
insert decode(curr) at the end of children of node
insert decode(curr) at the end of children of node
curr := right of curr
curr := right of curr
return node
return node
Let us see the following implementation to get a better understanding −
 Live Demo
#include <bits/stdc++.h>
using namespace std;
class TreeNode {
public:
   int val;
   TreeNode *left, *right;
   TreeNode(int data) {
      val = data;
      left = NULL;
      right = NULL;
   }
};
void inord(TreeNode *root) {
   if (root != NULL) {
      inord(root->left);
      cout << root->val << " ";
      inord(root->right);
   }
}
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:
   TreeNode* encode(Node* root) {
      if(!root) return NULL;
         TreeNode* node = new TreeNode(root->val);
         if(root->children.size()){
            node->left = encode(root->children[0]);
         }
         TreeNode* curr = node->left;
         for(int i = 1; i < root->children.size(); i++){
            curr->right = encode(root->children[i]);
            curr = curr->right;
         }
         return node;
      }
      Node* decode(TreeNode* root) {
         if(!root) return NULL;
            Node* node = new Node(root->val);
         TreeNode* curr = root->left;
         while(curr){
            node->children.push_back(decode(curr));
            curr = curr->right;
         }
      return node;
   }
};
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;
   cout << "Serialized Binary Tree: ";
   TreeNode *root = ob.encode(&n1);
   inord(root);
   cout << endl;
   Node *deser = ob.decode(root);
   cout << "Deserialized Tree: " << n_ary_to_str(deser);
}
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);
Given Tree: 1[3[5, 6, ], 2, 4, ]
Serialized Binary Tree: 5 6 3 2 4 1
Deserialized Tree: 1[3[5, 6, ], 2, 4, ]