Advanced Search

Posted By

marmomcil on 01/06/11

Tagged

Versions (?)

Last Edited at 01/06/11 07:21am

Statistics

Viewed 612 times
Favorited by 0 user(s)

Related snippets

668
bstablo_polje.h
627
bstablo_polje.h
603
bstablo_polje.h
600
zaglavlje bstablo_polje.h
617
bstablo_polje.h
489
bstablo_pokazivac.h - marmomcil
618
bstablo_polje.h
595
bstablo_polje.h
653
bstablo_polje.h

bstablo_polje.h - marmomcil

/ Published in: C++
Save to your folder(s)
Header file bstablo_polje.h - odn. implementacija binarnog stabla pomoću polja

Expand | Embed | Plain Text

Copy this code and paste it in your HTML 
  1. struct element{
  2. 	labeltype oznaka;
  3. 	int zapisan;
  4. }; 
  5.  
  6. struct tstablo{
  7. 	struct element polje[10000];
  8. };
  9.  
  10. typedef struct tstablo *tree;
  11. typedef int node;
  12.  
  13. node ParentB(node n,tree stablo){
  14. 	if(n==1) return -1;	
  15. 	else return ((int)n/2);
  16. }
  17.  
  18. node LeftChildB(node n,tree stablo){
  19. 	if (stablo->polje[2*n].zapisan) return 2*n;
  20. 	else return -1;
  21. }
  22.  
  23. node RightChildB(node n,tree stablo){
  24. 	if (stablo->polje[2*n+1].zapisan) return(2*n+1);
  25. 	else return -1;
  26. }
  27.  
  28. labeltype LabelB(node n,tree stablo){
  29. 	return(stablo->polje[n].oznaka);
  30. }
  31.  
  32. void ChangeLabelB(labeltype x, node n,tree stablo){
  33. 	if (stablo->polje[n].zapisan) stablo->polje[n].oznaka = x;
  34. 	else return;
  35. 	}
  36. }
  37.  
  38. node RootB(tree stablo){
  39. 	return 1;
  40. }
  41.  
  42. void CreateLeftB(labeltype x,node n,tree stablo){
  43. 	if (stablo->polje[2*n].zapisan) return;
  44. 	else{
  45. 		stablo->polje[2*n].zapisan = -1;
  46. 		stablo->polje[2*n].oznaka = x;
  47. 	}
  48. }
  49.  
  50. void CreateRightB(labeltype x,node n,tree stablo){
  51. 	if (stablo->polje[2*n+1].zapisan) return;
  52. 	else{
  53. 		stablo->polje[2*n+1].zapisan = -1;
  54. 		stablo->polje[2*n+1].oznaka = x;
  55. 	}
  56. }
  57.  
  58. void DeleteB(node n,tree stablo){
  59. 	node i;
  60. 	stack S;
  61. 	if(!stablo->polje[n].zapisan) return;	         
  62. 	else{   					
  63. 	  MakeNullS(&S);   				
  64. 	  if (stablo->polje[2*n].zapisan) PushS(2*n,&S);        
  65. 	  if (stablo->polje[2*n+1].zapisan) PushS(2*n+1,&S);          
  66. 	  stablo->polje[n].zapisan=0;  
  67. 	  while(!IsEmptyS(&S)){   			
  68. 	        i=TopS(&S);  				
  69.             PopS(&S);  				
  70.             if (stablo->polje[2*i].zapisan) PushS(2*i,&S);   
  71.             if (stablo->polje[2*i+1].zapisan) PushS(2*i+1,&S);     
  72. 	        stablo->polje[i].zapisan=0;  			
  73.     }
  74. }
  75.  
  76. void InitB(tree stablo, node x){
  77. 	for (int i=0;i<=9999;i++) stablo->polje[i].zapisan=-1;
  78. 	stablo->polje[1].oznaka=x;
  79. 	stablo->polje[1].zapisan=-1;
  80. }

Report this snippet


Comments

RSS Icon Subscribe to comments

You need to login to post a comment.