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Submitted By kloiz

Words 1523

Pages 7

Words 1523

Pages 7

Number Of Nodes:

1. /* 2. * C Program to Find the Number of Nodes in a Binary Tree 3. */ 4. #include <stdio.h> 5. #include <stdlib.h> 6. 7. /* 8. * Structure of node 9. */ 10. struct btnode 11. { 12. int value; 13. struct btnode *l; 14. struct btnode *r; 15. }; 16. 17. void createbinary(); 18. void preorder(node *); 19. int count(node*); 20. node* add(int); 21. 22. typedef struct btnode node; 23. node *ptr, *root = NULL; 24. 25. int main() 26. { 27. int c; 28. 29. createbinary(); 30. preorder(root); 31. c = count(root); 32. printf("\nNumber of nodes in binary tree are:%d\n", c); 33. } 34. /* 35. * constructing the following binary tree 36. * 50 37. * / \ 38. * 20 30 39. * / \ 40. * 70 80 41. * / \ \ 42. *10 40 60 43. */ 44. void createbinary() 45. { 46. root = add(50); 47. root->l = add(20); 48. root->r = add(30); 49. root->l->l = add(70); 50. root->l->r = add(80); 51. root->l->l->l = add(10); 52. root->l->l->r = add(40); 53. root->l->r->r = add(60); 54. } 55. 56. /* 57. * Add the node to binary tree 58. */ 59. node* add(int val) 60. { 61. ptr = (node*)malloc(sizeof(node)); 62. if (ptr == NULL) 63. { 64. printf("Memory was not allocated"); 65. return; 66. } 67. ptr->value = val; 68. ptr->l = NULL; 69. ptr->r = NULL; 70. return ptr; 71. } 72. 73. /* 74. * counting the number of nodes in a tree 75. */ 76. int count(node *n) 77. { 78. int c = 1; 79. 80. if (n == NULL) 81. return 0; 82. else 83. { 84. c += count(n->l); 85. c += count(n->r); 86. return c; 87. } 88. } 89. 90. /* 91. * Displaying the nodes of tree in preorder 92. */ 93. void preorder(node *t) 94. { 95. if (t != NULL) 96. { 97. printf("%d->", t->value); 98. preorder(t->l); 99. preorder(t->r); 100. } 101. }

No. Of Leaf Nodes:

#include <stdio.h>

#include <stdlib.h> /* A binary tree node has data, pointer to left child and a pointer to right child */ struct node

{

int data; struct node* left; struct node* right;

};

/* Function to get the count of leaf nodes in a binary tree*/ unsigned int getLeafCount(struct node* node)

{

if(node == NULL) return 0; if(node->left == NULL && node->right==NULL) return 1; else return getLeafCount(node->left)+ getLeafCount(node->right);

}

/* Helper function that allocates a new node with the given data and NULL left and right pointers. */ struct node* newNode(int data)

{

struct node* node = (struct node*) malloc(sizeof(struct node)); node->data = data; node->left = NULL; node->right = NULL; return(node);

}

/*Driver program to test above functions*/ int main()

{

/*create a tree*/ struct node *root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); /*get leaf count of the above created tree*/ printf("Leaf count of the tree is %d", getLeafCount(root)); getchar(); return 0;

}

Depth Of The Tree:

#include<stdio.h>

#include<stdlib.h>

/* A binary tree node has data, pointer to left child and a pointer to right child */ struct node

{

int data; struct node* left; struct node* right;

};

/* Compute the "maxDepth" of a tree -- the number of nodes along the longest path from the root node down to the farthest leaf node.*/ int maxDepth(struct node* node)

{

if (node==NULL) return 0; else { /* compute the depth of each subtree */ int lDepth = maxDepth(node->left); int rDepth = maxDepth(node->right); /* use the larger one */ if (lDepth > rDepth) return(lDepth+1); else return(rDepth+1); }

}

/* Helper function that allocates a new node with the given data and NULL left and right pointers. */ struct node* newNode(int data)

{

struct node* node = (struct node*) malloc(sizeof(struct node)); node->data = data; node->left = NULL; node->right = NULL; return(node);

}

int main()

{

struct node *root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); printf("Hight of tree is %d", maxDepth(root)); getchar(); return 0;

}

Number of internal Nodes = no of nodes- no of leaf nodes

No. of internal nodes =

1. /*

2. * C Program to Find the Number of Nodes in a Binary Tree

3. */

4. #include <stdio.h>

5. #include <stdlib.h>

6.

7. /*

8. * Structure of node

9. */

10. struct btnode

11. {

12. int value;

13. struct btnode *l;

14. struct btnode *r;

15. };

16.

17. void createbinary();

18. void preorder(node *);

19. int count(node*);

20. node* add(int);

21.

22. typedef struct btnode node;

23. node *ptr, *root = NULL; unsigned int getLeafCount(struct node* node)

{

if(node == NULL) return 0; if(node->left == NULL && node->right==NULL) return 1; else return getLeafCount(node->left)+ getLeafCount(node->right);

}

25. int main()

26. {

27. int c;

28.

29. createbinary();

30. preorder(root);

31. c = count(root); int g = getLeafCount(root); int internal = c- g;

32. printf("\nNumber of internal nodes in binary tree are:%d\n", internal);

33. }

34. /*

35. * constructing the following binary tree

36. * 50

37. * / \

38. * 20 30

39. * / \

40. * 70 80

41. * / \ \

42. *10 40 60

43. */

44. void createbinary()

45. {

46. root = add(50);

47. root->l = add(20);

48. root->r = add(30);

49. root->l->l = add(70);

50. root->l->r = add(80);

51. root->l->l->l = add(10);

52. root->l->l->r = add(40);

53. root->l->r->r = add(60);

54. }

55.

56. /*

57. * Add the node to binary tree

58. */

59. node* add(int val)

60. {

61. ptr = (node*)malloc(sizeof(node));

62. if (ptr == NULL)

63. {

64. printf("Memory was not allocated");

65. return;

66. }

67. ptr->value = val;

68. ptr->l = NULL;

69. ptr->r = NULL;

70. return ptr;

71. }

72.

73. /*

74. * counting the number of nodes in a tree

75. */

76. int count(node *n)

77. {

78. int c = 1;

79.

80. if (n == NULL)

81. return 0;

82. else

83. {

84. c += count(n->l);

85. c += count(n->r);

86. return c;

87. }

88. }

89.

90. /*

91. * Displaying the nodes of tree in preorder

92. */

93. void preorder(node *t)

94. {

95. if (t != NULL)

96. {

97. printf("%d->", t->value);

98. preorder(t->l);

99. preorder(t->r);

100. }

101. }

For 1 through 6, the most important concept is of the Big-Oh notation f(n) = O(g(n)) means there are positive constants c and k, such that 0 ≤ f(n) ≤ cg(n) for all n ≥ k. The values of c and k must be fixed for the function f and must not depend on n.

So, for the questions

Since We have to prove that f(n) < cg(n), only the terms with the HIGHEST COEFFECIENT matters, or the term which has the MAXIMUM VALUE

1. The maximum of those two will be used for the Big-Oh definition. This is just the definition of Big-Oh.

Since O(f(n)+g(n)) is effectively the O() of whichever one if bigger, that is exactly what’s written on the Left hand side.

2. Similarly, since n^5 is the fastest growing term, this is the only term which will matter in finding the Big-Oh, since big Oh needs to prove that f(n) can be SUPPRESSED by g(n) in every case

3. 2^n+1 can be written as 2*2^n, which is of the form c*2^n. Therefore, 2^n+1 = O(2^n)

4. Sum of i^2, i=0 to n is equal to n*(n+1)*(2n+1)/6 which in turn has n^3 as the highest degree term, i.e. the fastest growing term. Hence, O(n^3) is the answer to this.

5. Use this as a practice exercise for yourself. It’s really easy and all you have to do is arrange them in their order of growth, i.e. loglogn grows slower than log^2 (n) , etc

2^100, loglogn, log^2n, sqrt(logn), 1/n, n ^0.01 ,[3sqrt(n), Sqrt(n)], 5n, [nlog4n,6nlogn], 2nlog^2n,4*n^3/2, n^2logn

2^logn, 4^logn, 2^n, [4^n, 2^2^n]

6. Wrong Question. T(n) will be equal to 2^(n+1) – 3

7. Root- if(p->parent == NULL)

Parent- if(p->leftChild !=NULL || p->rightChild !=NULL)

Left Child if(p->parent->left == p)

Right Child id(p->parent->right == p)

isInternal if(p->leftChild !=NULL &&p->rightChild !=NULL && p->parent!=NULL)

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