c语言实验-树解读.doc

1、验证基于顺序存储的二叉树的基本操作。 #include iostream.h #define NULL 0 #define MaxSize 100 typedef struct sqtree { char data[MaxSize]; int num; }SqTree; void InitTree(SqTree *t) { t = new SqTree; t-num = 0; } void CreaTree(SqTree *t, char tr[]) { for (int i = 0; tr[i] != \0; i ++) { t-data[i] = tr[i]; t-num ++; } } void DispTree(SqTree *t) { for (int i = 0; i t-num; i ++) { coutt-data[i]; } coutendl; } void FindTNode(SqTree *t, char x) { for (int i = 0; i t-num; i ++) { if (x == t-data[i]) { if (i = t-num/2) { cout值为x的结点为叶子结点,其双亲为:; coutt-data[(i+1)/2]endl; } else if (i == 0) { cout值为x的结点为根结点endl; cout值为x的结点的左孩子结点值为:; coutt-data[2*i+1]endl; cout值为x的结点的右孩子结点值为:; coutt-data[2*i+2]endl; } else { cout值为x的结点的双亲结点值为:; coutt-data[(i+1)/2]endl; cout值为x的结点的左孩子结点值为:; coutt-data[2*i+1]endl; cout值为x的结点的右孩子结点值为:; coutt-data[2*i+2]endl; } return; } } cout值为x的节点不存在!endl; } int main() { SqTree *T= NULL; char a[] = abcdefghijklmn; char x; InitTree(T); CreaTree(T, a); DispTree(T); cout请输入要查找的节点:; cinx; FindTNode(T, x); return 0; } 2、验证基于链式存储的二叉树的基本操作。 3、已知某电文中仅有8个不同的字符，各字符出现的次数依次是3，4，8，10，16，18，20，21，请设计程序，实现对各个字符的哈夫曼编码。 #include iostream.h #define MaxSize 50 typedef struct { char data; int weight; int parent; int lchild; int rchild; } HTNode; typedef struct { char cd[MaxSize]; int start; } HCode; void CreateHT(HTNode ht[], int n) { int i, k, lnode, rnode; float min1, min2; for (i = 0; i 2*n-1; i++) ht[i].parent = ht[i].lchild = ht[i].rchild = -1; for (i = n; i 2*n-1; i++) { min1 = min2 = 32767; lnode = rnode = -1; for (k = 0; k i; k++) { if (ht[k].parent == -1) { if (ht[k].weight min1) { min2 = min1; rnode = lnode; min1 = ht[k].weight; lnode = k; } else if (ht[k].weight min2) { min2 = ht[k].weight; rnode = k