Tree: Huffman Decoding

Problem Statement :

Huffman coding assigns variable length codewords to fixed length input characters based on their frequencies. More frequent characters are assigned shorter codewords and less frequent characters are assigned longer codewords. All edges along the path to a character contain a code digit. If they are on the left side of the tree, they will be a 0 (zero). If on the right, they'll be a 1 (one). Only the leaves will contain a letter and its frequency count. All other nodes will contain a null instead of a character, and the count of the frequency of all of it and its descendant characters.

For instance, consider the string ABRACADABRA. There are a total of 11 characters in the string. This number should match the count in the ultimately determined root of the tree. Our frequencies are A = 5 , B =2, R = 2, C = 1 and D = 1 . The two smallest frequencies are for C and D , both equal to 1 , so we'll create a tree with them. The root node will contain the sum of the counts of its descendants, in this case 1 + 1  = 2. The left node will be the first character encountered, C  , and the right will contain D. Next we have 3 items with a character count of 2: the tree we just created, the character B and the character R. The tree came first, so it will go on the left of our new root node. B will go on the right. Repeat until the tree is complete, then fill in the 1's and 0's for the edges. The finished graph looks like:

Input characters are only present in the leaves. Internal nodes have a character value of ϕ (NULL). We can determine that our values for characters are:

A - 0
B - 111
C - 1100
D - 1101
R - 10
Our Huffman encoded string is:

A B    R  A C     A D     A B    R  A
0 111 10 0 1100 0 1101 0 111 10 0
To avoid ambiguity, Huffman encoding is a prefix free encoding technique. No codeword appears as a prefix of any other codeword.

To decode the encoded string, follow the zeros and ones to a leaf and return the character there.

You are given pointer to the root of the Huffman tree and a binary coded string to decode. You need to print the decoded string.

Function Description

Complete the function decode_huff in the editor below. It must return the decoded string.

decode_huff has the following parameters:

root: a reference to the root node of the Huffman tree
s: a Huffman encoded string
Input Format

There is one line of input containing the plain string, . Background code creates the Huffman tree then passes the head node and the encoded string to the function.

   1 <=  |s| <= 25

Output Format

Output the decoded string on a single line.

Solution :


                            Solution in C :

In C++ :

The structure of the node is

typedef struct node
    int freq;
    char data;
    node * left;
    node * right;


void decode_huff(node * root,string s)
   node *treeroot = root;
  // char *result = (char*)malloc(sizeof(char)*len);
    int i=0,k=0;
   while (s[i]!='\0'){
    //   printf("s[%d]%c\n", i, s[i]);
       if (s[i]== '1' ){
         //  printf("i was here");
           if (root->data != '\0'){
           //    printf("i was here");
                 printf("%c", root->data);
                 root = treeroot;
       else {
            if (root->data != '\0'){
                 printf("%c", root->data);
                root = treeroot;

In Java :

   class Node
      public  int frequency; // the frequency of this tree
       public  char data;
       public  Node left, right;

void decode(String S ,Node root)
        Node temp=root;
        String ans="";
        for(int i=0;i<S.length();i++){
         //   System.out.println("er1");
            if(temp.right==null && temp.left==null)

In python3 :

"""class Node:
    def __init__(self, freq,data):
        self.freq= freq
        self.left = None
        self.right = None
def decodeHuff(root, s):
    s=[int(i) for i in s]
    while len(s)!=0 :

# Enter your code here. Read input from STDIN. Print output to STDOUT
def decodeHuf(root, s,parent):
    if not (root.left or root.right) :
    if len(s)==0 :
    del s[0]
    if x == 1 :
    else :
	#Enter Your Code Here

View More Similar Problems

Game of Two Stacks

Alexa has two stacks of non-negative integers, stack A = [a0, a1, . . . , an-1 ] and stack B = [b0, b1, . . . , b m-1] where index 0 denotes the top of the stack. Alexa challenges Nick to play the following game: In each move, Nick can remove one integer from the top of either stack A or stack B. Nick keeps a running sum of the integers he removes from the two stacks. Nick is disqualified f

View Solution →

Largest Rectangle

Skyline Real Estate Developers is planning to demolish a number of old, unoccupied buildings and construct a shopping mall in their place. Your task is to find the largest solid area in which the mall can be constructed. There are a number of buildings in a certain two-dimensional landscape. Each building has a height, given by . If you join adjacent buildings, they will form a solid rectangle

View Solution →

Simple Text Editor

In this challenge, you must implement a simple text editor. Initially, your editor contains an empty string, S. You must perform Q operations of the following 4 types: 1. append(W) - Append W string to the end of S. 2 . delete( k ) - Delete the last k characters of S. 3 .print( k ) - Print the kth character of S. 4 . undo( ) - Undo the last (not previously undone) operation of type 1 or 2,

View Solution →

Poisonous Plants

There are a number of plants in a garden. Each of the plants has been treated with some amount of pesticide. After each day, if any plant has more pesticide than the plant on its left, being weaker than the left one, it dies. You are given the initial values of the pesticide in each of the plants. Determine the number of days after which no plant dies, i.e. the time after which there is no plan

View Solution →

AND xor OR

Given an array of distinct elements. Let and be the smallest and the next smallest element in the interval where . . where , are the bitwise operators , and respectively. Your task is to find the maximum possible value of . Input Format First line contains integer N. Second line contains N integers, representing elements of the array A[] . Output Format Print the value

View Solution →


You are a waiter at a party. There is a pile of numbered plates. Create an empty answers array. At each iteration, i, remove each plate from the top of the stack in order. Determine if the number on the plate is evenly divisible ith the prime number. If it is, stack it in pile Bi. Otherwise, stack it in stack Ai. Store the values Bi in from top to bottom in answers. In the next iteration, do the

View Solution →