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  1. Tree: Huffman Decoding
  2. Discussions

Tree: Huffman Decoding

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80 Discussions

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  • anarod_val25
     + 0 comments
    void iter(node * root, std::string key, std::map<std::string, std::string> &dictionary){

if(root == NULL){
    return;
}

if(root->data != '\0') {
    std::string s{root->data};
    dictionary[key] = s;
}

iter(root->left, key+"0", dictionary);
iter(root->right, key+"1", dictionary);
}

void decode_huff(node * root, string s) {
std::map<std::string, std::string> dictionary;   
std::string binary = "";

iter(root, binary, dictionary);

binary="";

for (char c : s){
    std::string str{c};
    binary = binary+str;

    if (dictionary.find(binary) != dictionary.end()) {
        std::cout <<dictionary[binary];
        binary="";
    }
}
}

  • tgfelchlin
     + 0 comments
    void decode(String s, Node root) {
        String word = "";
        Node location = root;
        for(char i: s.toCharArray()){
       if(i == '0'){
        if (location.left.left ==null){
            word+=location.left.data;
            location = root;
        }else{
        location = location.left;
        }
       }
       else if(i=='1'){
        if (location.right.right ==null){
            word+=location.right.data;
            location = root;
        }else{
        location = location.right;
        }
       }
       } 
       System.out.print(word);

  • yschoss
     + 0 comments

    Python coders beware: when node.data does not contain a character, it is not an empty string or None. It is '\0'.

    This drove me nuts for 20 min until I read some discussions.

    I feel like this was a poor choice of "empty" string. Why not just set it to None or ""? I guess I learned something new about Python today but it had nothing to do with trees. Anyone disagree?

  • uchihanigobi
     + 0 comments
    def decodeHuff(root, s):
	#Enter Your Code Here
    s = list(s)
    current = root
    results = []
    
    def traverse(current, s):
        if len(s) > 0:
            if s[0] == "0" and current.left is not None:
                s.pop(0)
                traverse(current.left, s)
            elif s[0] == "1" and current.right is not None:
                s.pop(0)
                traverse(current.right, s)
        
        if current.left is None and current.right is None:
            results.append(current.data)

    while len(s) > 0:
        traverse(current, s)
    
    print("".join(results))

  • wasifk308
     + 0 comments
    void PrintCode(node* node , string s , int index  , struct node* root)
{
    if(node == nullptr)
    {
        return;
    }
     index = index+1;
    if(index >s.length())
    {
        return ;
    }
   

   // cout<<node->freq;
    //Leaf node reached
    if(node->left == nullptr && node->right == nullptr )
    { 
       cout<<node->data;
        node = root;
    }
    
    //Move left
    if(s[index] == '0')
    { 
        PrintCode(node->left , s, index , root);

    }else if(s[index] == '1')
    {
        PrintCode(node->right , s, index , root);
    }
   
}


void PrintLeafNode(node * node)
{
    
    if(node == nullptr)
    {
        return;
    }
    if(node->left == nullptr && node->right == nullptr)
    {    
        cout<<node->data  << " "<< node->freq << endl;
    }
    PrintLeafNode(node->left);
    PrintLeafNode(node->right);
}
void decode_huff(node * root, string s) {

    //given is root node // we can traverse root node  based on the value s
   PrintCode(root,s,-1,root);
    
}