Lecture27ECE264Spring2012 - Rhea

Hanye Xu April 20 Lecture 27

264 exam outcome 1.file 2.structure 3.dynamic structure (linked list)

binary search tree

typedef struct treenode { struct treenode *left; struct treenode *right; int value; }Node;

int Tree_search(Node *n, int v) { /*return 0 if not found,1 if found*/ if(n == NULL){return 0;} if((n->value)==v){return 1;} if((n->value)>v) { return Tree_search(n->left,v); } return Tree_search(n->right,v); }

void Tree_destroy(Node *n) { if(n == NULL){return;} Tree_destroy(n->left); Tree_destroy(n->right); free(n); }

void Tree_print(Node *n) { if(n==NULL){return;} (1)printf("%d",n->value); 1 2 3 -> preorder 6 2 0 4 9 7 (2)Tree_print(n->left); 2 1 3 -> inorder sorting 0 2 4 6 7 9 (3)Tree_print(n->right); 2 3 1 -> postorder 0 4 2 7 9 6 }

Node *Node_construct(int v) { Node *n; n = malloc(sizeof(Node)); n->value = v; n->left = NULL; n->right = NULL; return n; }

Node *Tree_construct(Node *n, int v) { if(n==NULL){return Node_construct(v);} if((n->value)==v){return n;} if((n->value)>v) { n->left = Node_construct(n->left,v); } else { n->right= Node_construct(n->right,v); } return n; }

insert 6,2,4,0,9,7 Node *n =NULL; n=Tree_insert(n,6);

n-> 6 (root) / \

      GRD

n=Tree_insert(n,2);

n-> 6 / \ 2 GRD

       /  \
      GRD

n=Tree_insert(n,4)

n-> 6

       /    \
    2      GRD
   /   \

GRD 4 / \ GRD

Finally n-> 6

                     /    \
                  2      9
                /   \     /  \
              0     4 7   GRD

parser

int y = 3+2; int t = 3 + "ece";

James Chen Notes

Binary search tree

A binary search tree has an initial root and left and right subtrees. Each subtree starts with a node and has left and right children It is possible that the left or right may be NULL if a node has a value v: everything in the left subtree < v everything in the right subtree > v

notes: insert 6, 2, 4 Node *n = NULL; n=Tree_insert(n, 6);//creates root of 6, its left and right subtrees are NULL n=Tree_insert(n, 2); //creates node with value 2, since it is < 6 and the left child of 6 is NULL, 2 becomes the left child n=Tree_insert(n,4); //value 4, it is <6 but, the left child is not NULL, the insert function is called using the left child (2). //Since 4 > 2, and the right child of 2 is NULL, 4 becomes the right child

each search checks value against node value, and moves left and right as appropriate binary search tree works best when balanced

typedef struct treenode { struct treenode *left; struct treenode *right; int value; }Node;

int Tree_search(Node* n, int v) // or Node* Tree_search, returns n or NULL { /*return 0 if not found, 1 if found */ if (n==NULL) {return 0;} if ((n->value) == v) {return 1;} if ((n->value) > v) { return Tree_search(n->left,v); } return Tree_search(n->right,v); }

void Tree_destroy(Node *n) { if(n==NULL) {return;} Tree_destroy(n->left); // destroy left child Tree_destroy(n->right); // destroy right child free(n); // destroy itself }

void Tree_print(Node *n) // prints least to greatest { if(n==NULL) {return;} printf(“%d”,n->value); // (1) Tree_print(n->left);// (2) Tree_print(n->right);// (3) }

/* sample tree; 6 2 9 0 4 7 1, 2, 3 → preorder 2, 1, 3 → in order // prints tree in order of least to greatest ( 0 2 4 6 7 9 ) 3, 1, 2 → for greatest to least ( 9 7 6 4 2 0 ) 2, 3, 1 → post order (hierarchy operation; like PEMDAS)

parser → the very first thing a compiler does. Analyze source code, breaks into smaller units, decides if the units can be put together

• /

Node* Node_construct(int v) { Node *n; n=malloc(sizeof(Node)); n->value = v; n->left = NULL; n->right = NULL; return n; }

Node *Tree_construct(Node *n, int v) { if(n==NULL) { return Node_construct(v);} } if((n->value) == v) { return n; } if((n->value)>n) { n->left = Node_construct(n->left,n); } else { n->right = Node_construct(n->right, v); } return n; }