Data Structure --- implementation of the Harman tree and coding in c language --- c fuman

Data Structure --- implementation of the Harman tree and coding in c language --- c fuman

// Harman tree // Yang Xin # include <stdio. h> # include <stdlib. h> typedef int ElemType; struct BTreeNode {ElemType data; struct BTreeNode * left; struct BTreeNode * right;}; // traverse the Harman tree void PrintBTree_int (struct BTreeNode * BT) {if (BT! = NULL) {printf ("% d", BT-> data); if (BT-> left! = NULL | BT-> right! = NULL) {printf ("("); PrintBTree_int (BT-> left); // output the left subtree if (BT-> right! = NULL) printf (","); PrintBTree_int (BT-> right); // outputs the right subtree printf (")");}}} // create a user tree struct BTreeNode * CreateHuffman (ElemType a [], int n) {int I, j; struct BTreeNode ** B, * q; B = malloc (n * sizeof (struct BTreeNode); for (I = 0; I <n; I ++) // dynamic memory allocation {B [I] = malloc (sizeof (struct BTreeNode); B [I]-> data = a [I]; B [I]-> left = B [I]-> right = NULL;} for (I = 1; I <n; I ++) {// k1 indicates the subscript of the root node with the minimum weight in the forest, and k2 indicates the minimum weight. Subscript int k1 =-1, k2; for (j = 0; j <n; j ++) // Let k1 point to the first tree in the forest, k2 points to the second {if (B [j]! = NULL & k1 =-1) {k1 = j; continue;} if (B [j]! = NULL) {k2 = j; break ;}for (j = k2; j <n; j ++) // construct the optimal solution {if (B [j]! = NULL) {if (B [j]-> data <B [k1]-> data) {k2 = k1; k1 = j ;} else if (B [j]-> data <B [k2]-> data) k2 = j;} q = malloc (sizeof (struct BTreeNode )); q-> data = B [k1]-> data + B [k2]-> data; q-> left = B [k1]; q-> right = B [k2]; B [k1] = q; B [k2] = NULL;} free (B); return q ;} // calculate the permission path ElemType WeightPathLength (struct BTreeNode * FBT, int len) // The initial len value is 0 {if (FBT = NULL) // return 0 return 0 for the empty tree; else {if (FBT-> left = N ULL & FBT-> right = NULL) return FBT-> data * len; else return WeightPathLength (FBT-> left, len + 1) + WeightPathLength (FBT-> right, len + 1) ;}}// construct the user code void HuffManCoding (struct BTreeNode * FBT, int len) {static int a [10]; if (FBT! = NULL) {if (FBT-> left = NULL & FBT-> right = NULL) {int I; printf ("encode the node value as % d: ", FBT-> data); for (I = 0; I <len; I ++) printf (" % d ", a [I]); printf ("\ n");} else {a [len] = 0; HuffManCoding (FBT-> left, len + 1); a [len] = 1; huffManCoding (FBT-> right, len + 1) ;}} int main () {int n, I; ElemType * a; struct BTreeNode * fbt; printf ("Enter the number of weighted nodes n: \ n"); while (1) {scanf ("% d", & n); if (n> 1) break; else printf ("re-input n value:");} a = malloc (n * sizeof (ElemType); printf ("Enter % d integers as the weight: \ n ", n); for (I = 0; I <n; I ++) scanf (" % d ", & a [I]); fbt = CreateHuffman (a, n); printf ("the husky tree is as follows: \ n"); PrintBTree_int (fbt); printf ("\ n "); printf ("length of the weighted path of the Harman tree: \ n"); printf ("% d \ n", WeightPathLength (fbt, 0 )); printf ("each leaf node in the tree is encoded as \ n"); HuffManCoding (fbt, 0); return 0 ;}

Copyright Disclaimer: This article is an original article by the blogger and cannot be reproduced without the permission of the blogger.