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

C++Server Side ProgrammingProgramming

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,

  • if root is valid, then −

    • return null

  • 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)

  • 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

  • return node

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

  • if root is not present, then −

    • return NULL

  • node := new node with val of root

  • curr := left of root

  • while curr is non-zero, do −

    • insert decode(curr) at the end of children of node

    • curr := right of curr

  • return node

Example 

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);
}

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, ]
Serialized Binary Tree: 5 6 3 2 4 1
Deserialized Tree: 1[3[5, 6, ], 2, 4, ]
raja
Published on 21-Jul-2020 12:26:48
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