Lecture 27 - Rhea

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";