Suppose we have a binary tree. As we know the succinct encoding of Binary Tree performs close to lowest possible space. The n’th Catalan number is designated by the number of structurally different binary trees with n different nodes. If the n is large, this is about 4n; thus, we require minimum about log2(4) n = 2n bits to encode it. A succinct binary tree therefore would consume 2n + O(n) bits.
So, if the input is like
then the output will be
encoded −
Structure List 1 1 1 0 0 1 0 0 1 0 1 0 0
Data List 10 20 40 50 30 70
Decoded − The tree as shown above.
To solve this, we will follow these steps −
Let us see the following implementation to get better understanding −
#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 Encode(TreeNode *root, list<bool>&struc, list<int>&data){ if(root == NULL){ struc.push_back(0); return; } struc.push_back(1); data.push_back(root->val); Encode(root->left, struc, data); Encode(root->right, struc, data); } TreeNode *Decode(list<bool>&struc, list<int>&data){ if(struc.size() <= 0) return NULL; bool b = struc.front(); struc.pop_front(); if(b == 1){ int key = data.front(); data.pop_front(); TreeNode *root = new TreeNode(key); root->left = Decode(struc, data); root->right = Decode(struc, data); return root; } return NULL; } void preorder_trav(TreeNode* root){ if(root){ cout << "key: "<< root->val; if(root->left) cout << " | left child: "<< root->left->val; if(root->right) cout << " | right child: "<< root->right->val; cout << endl; preorder_trav(root->left); preorder_trav(root->right); } } main() { TreeNode *root = new TreeNode(10); root->left = new TreeNode(20); root->right = new TreeNode(30); root->left->left = new TreeNode(40); root->left->right = new TreeNode(50); root->right->right = new TreeNode(70); cout << "The Tree\n"; preorder_trav(root); list<bool> struc; list<int> data; Encode(root, struc, data); cout << "\nEncoded Tree\n"; cout << "Structure List\n"; list<bool>::iterator si; // Structure iterator for(si = struc.begin(); si != struc.end(); ++si) cout << *si << " "; cout << "\nData List\n"; list<int>::iterator di; // Data iIterator for(di = data.begin(); di != data.end(); ++di) cout << *di << " "; TreeNode *newroot = Decode(struc, data); cout << "\n\nPreorder traversal of decoded tree\n"; preorder_trav(newroot); }
root->left = new TreeNode(20); root->right = new TreeNode(30); root->left->left = new TreeNode(40); root->left->right = new TreeNode(50); root->right->right = new TreeNode(70);
The Tree key: 10 | left child: 20 | right child: 30 key: 20 | left child: 40 | right child: 50 key: 40 key: 50 key: 30 | right child: 70 key: 70 Encoded Tree Structure List 1 1 1 0 0 1 0 0 1 0 1 0 0 Data List 10 20 40 50 30 70 Preorder traversal of decoded tree key: 10 | left child: 20 | right child: 30 key: 20 | left child: 40 | right child: 50 key: 40 key: 50 key: 30 | right child: 70 key: 70