我们如何找到以下代码中找到B树中的级别数
#include<stdio.h>
#include<stdlib.h>
#define M 10
struct node
{
int n; /* n < M No. of keys in node will always less than order of Btree */
int keys[M-1]; /*array of keys*/
struct node *p[M]; /* (n+1 pointers will be in use) */
}*root=NULL;
enum KeyStatus { Duplicate,SearchFailure,Success,InsertIt,LessKeys };
void insert(int key);
void display(struct node *root,int);
enum KeyStatus ins(struct node *r, int x, int* y, struct node** u);
int searchPos(int x,int *key_arr, int n);
int main()
{
int key;
int choice;
int i,j,mul,count=0;
int mul_count[65026]={0},value[7097];
for(i=1;i<=255;i++)
{
for(j=1;j<=255;j++)
{
mul = j*i;
mul_count[mul]++;
}
}
for(i=1;i<=65026;i++)
{
if( mul_count[i]!= 0 && mul_count[i]!= 2)
{
value[count]=i;
count++;
}
}
printf("Creation of B tree for node %d\n", M);
while(1)
{
printf("1.Insert\n");
printf("2.Display\n");
printf("3.Quit\n");
printf("Enter your choice : ");
scanf("%d",&choice);
switch(choice)
{
case 1:
printf("Inserting is done automatically press 2: \n");
//scanf("%d",&key);
for(i=0;i<count;i++)
{
key = value[i];
insert(key);
}
break;
case 2:
printf("Btree is :\n");
display(root,0);
break;
ase 3:
exit(1);
default:
printf("Wrong choice\n");
break;
}/*End of switch*/
}/*End of while*/
return 0;
}/*End of main()*/
void insert(int key)
{
struct node *newnode;
int upKey;
enum KeyStatus value;
value = ins(root, key, &upKey, &newnode);
if (value == Duplicate)
printf("Key already available\n");
if (value == InsertIt)
{
struct node *uproot = root;
root=malloc(sizeof(struct node));
root->n = 1;
root->keys[0] = upKey;
root->p[0] = uproot;
root->p[1] = newnode;
}/*End of if */
}/*End of insert()*/
enum KeyStatus ins(struct node *ptr, int key, int *upKey,struct node **newnode)
{
struct node *newPtr, *lastPtr;
int pos, i, n,splitPos;
int newKey, lastKey;
enum KeyStatus value;
if (ptr == NULL)
{
*newnode = NULL;
*upKey = key;
return InsertIt;
}
n = ptr->n;
pos = searchPos(key, ptr->keys, n);
if (pos < n && key == ptr->keys[pos])
return Duplicate;
value = ins(ptr->p[pos], key, &newKey, &newPtr);
if (value != InsertIt)
return value;
/*If keys in node is less than M-1 where M is order of B tree*/
if (n < M - 1)
{
pos = searchPos(newKey, ptr->keys, n);
/*Shifting the key and pointer right for inserting the new key*/
for (i=n; i>pos; i--)
{
ptr->keys[i] = ptr->keys[i-1];
ptr->p[i+1] = ptr->p[i];
}
/*Key is inserted at exact location*/
ptr->keys[pos] = newKey;
ptr->p[pos+1] = newPtr;
++ptr->n; /*incrementing the number of keys in node*/
return Success;
}/*End of if */
/*If keys in nodes are maximum and position of node to be inserted is last*/
if (pos == M - 1)
{
lastKey = newKey;
lastPtr = newPtr;
}
else /*If keys in node are maximum and position of node to be inserted is not last*/
{
lastKey = ptr->keys[M-2];
lastPtr = ptr->p[M-1];
for (i=M-2; i>pos; i--)
{
ptr->keys[i] = ptr->keys[i-1];
ptr->p[i+1] = ptr->p[i];
}
ptr->keys[pos] = newKey;
ptr->p[pos+1] = newPtr;
}
splitPos = (M - 1)/2;
(*upKey) = ptr->keys[splitPos];
(*newnode)=malloc(sizeof(struct node));/*Right node after split*/
ptr->n = splitPos; /*No. of keys for left splitted node*/
(*newnode)->n = M-1-splitPos;/*No. of keys for right splitted node*/
for (i=0; i < (*newnode)->n; i++)
{
(*newnode)->p[i] = ptr->p[i + splitPos + 1];
if(i < (*newnode)->n - 1)
(*newnode)->keys[i] = ptr->keys[i + splitPos + 1];
else
(*newnode)->keys[i] = lastKey;
}
(*newnode)->p[(*newnode)->n] = lastPtr;
return InsertIt;
}/*End of ins()*/
void display(struct node *ptr, int blanks)
{
int check = 0;
if (ptr)
{
int i;
for(i=1;i<=blanks;i++)
printf(" ");
for (i=0; i < ptr->n; i++)
{
//check++;
printf("%d ",ptr->keys[i]);
}
printf("\n");
for (i=0; i <= ptr->n; i++)
display(ptr->p[i], blanks+10);
}/*End of if*/
//printf("\n no of levels:%d",check);
}/*End of display()*/
int searchPos(int key, int *key_arr, int n)
{
int pos=0;
while (pos < n && key > key_arr[pos])
pos++;
return pos;
}/*End of searchPos()*/
答案 0 :(得分:0)
有点像你的display
功能:
int max(int a, int b)
{
return a > b ? a : b;
}
int depth(struct node *ptr)
{
int d=0;
if (ptr==NULL)
return d;
for (i=0; i <= ptr->n; i++)
d=max(d,depth(ptr->p[i]))
return d+1;
}