int countTrees(int numKeys) {
int count=0;
if(numKeys == 1 || numKeys == 0)
return 1;
if(numKeys == 2)
return 2;
for(int i=1;i<=numKeys;i++) {
count += countTrees(numKeys - i) * countTrees(i - 1);
}
return count;
}
/*
* For the key values 1...numKeys,
* how many structurally unique binary search trees are possible
* that store those keys.
*
* Strategy: consider that each value could be the root.
*/
int countTrees(int numKeys){
if (numKeys <= 1){
return 1;
}else{
int sum = 0;
int left, right, root;
for (root = 1; root <= numkeys; ++root){
left = countTrees(root - 1);
right = countTrees(numKeys - root);
// number of possible trees with this root = left*right
sum += left*right;
}
return sum;
}
}
void CATrees() {
for (int y = 0; y < torch.buf.geth(); y++) {
for (int x = 0; x < torch.buf.getw(); x++) {
if (!game.map.solid(x,y) && (rand4() < 6)) set_tile(x,y, TREE);
}
}
bool *next = new bool[torch.buf.geth()*torch.buf.getw()];
for (int i = 0; i < 4; i++) {
memset(next, 0, torch.buf.geth()*torch.buf.getw()*sizeof(bool));
for (int y = 0; y < torch.buf.geth(); y++) {
for (int x = 0; x < torch.buf.getw(); x++) {
if (countTrees(x,y,2) <= 3 || countTrees(x,y,1) >= 5) next[y*torch.buf.getw()+x] = true;
else next[y*torch.buf.getw()+x] = false;
}
}
for (int y = 0; y < torch.buf.geth(); y++) {
for (int x = 0; x < torch.buf.getw(); x++) {
if (!game.map.solid(x,y) || game.map.at(x,y)->type == TREE) {
if (next[y*torch.buf.getw()+x])
set_tile(x,y, TREE);
else
set_tile(x,y, GROUND);
}
}
}
}
for (int i = 0; i < 3; i++) {
memset(next, 0, torch.buf.geth()*torch.buf.getw()*sizeof(bool));
for (int y = 0; y < torch.buf.geth(); y++) {
for (int x = 0; x < torch.buf.getw(); x++) {
if (countTrees(x,y,1) >= 5) next[y*torch.buf.getw()+x] = true;
else next[y*torch.buf.getw()+x] = false;
}
}
for (int y = 0; y < torch.buf.geth(); y++) {
for (int x = 0; x < torch.buf.getw(); x++) {
if (!game.map.solid(x,y) || game.map.at(x,y)->type == TREE) {
if (next[y*torch.buf.getw()+x])
set_tile(x,y, TREE);
else
set_tile(x,y, GROUND);
}
}
}
}
delete [] next;
}
int countTrees(int n){
if(n<=1){
return 1;
}
else{
int sum, left,right, root;
sum = 0;
for(root=1;root<=n;root++){
left = countTrees(root-1);
right =countTrees(n-root);
sum += left*right;
}
return sum;
}
}
int main()
{
// printf("Hello world!\n");
/*
4
/ \
2 8
/ \
5 10
\
7
/
6
*/
struct BST* root = newNode(4);
root->left = newNode(2);
root->right = newNode(8);
root->right->left = newNode(5);
root->right->right = newNode(10);
root->right->left->right = newNode(7);
root->right->left->right->left = newNode(6);
struct BST *temp;
int a = -1;
// int *p = &a;
/*
if(checkForBST(root,&a)){
printf("Tree is BST\n");
}
else{
printf("Tree is not BST\n");
}
a= 3;
temp = kthSmallestElement(root,&a);
printf("Kth Smallest Element : %d\n",temp->data);
temp = inOrderSuccessor(root,root->right->left->right);
printf("Inorder Successor of 7:: %d\n",temp->data);
deleteNode(root,8);
temp = inOrderSuccessor(root,root->right->left->right);
printf("Inorder Successor of 7:: %d\n",temp->data);
*/
printBST(root);
printf("\n");
temp = TrimBST(root,3,7);
//printBST(temp);
printf("\nNumber of Trees with 3 Nodes : %d\n",countTrees(3));
return 0;
}
/*
For the key values 1...numKeys, how many structurally unique
binary search trees are possible that store those keys.
Strategy: consider that each value could be the root.
Recursively find the size of the left and right subtrees.
*/
int countTrees(int numKeys) {
if (numKeys <=1) {
return(1);
}
else {
// there will be one value at the root, with whatever remains
// on the left and right each forming their own subtrees.
// Iterate through all the values that could be the root...
int sum = 0;
int left, right, root;
for (root=1; root<=numKeys; root++) {
left = countTrees(root - 1);
right = countTrees(numKeys - root);
// number of possible trees with this root == left*right
sum += left*right;
}
return(sum);
}
}
int main()
{
#if 0
int a[] = {2, 4, 1, 8, 3, 5, 9, 6, 7};
tree_t root = NULL;
for(int i = 0; i < 9; ++i)
{
root = insert(root,a[i]);
}
display(root);
printf("----------------------\n");
#endif
#if 0
printf("In-order traversal \n");
display(root);
printf("No of nodes: %d\n", count(root));
printf("----------------------\n");
printf("No of leaf nodes: %d\n",count_leaf(root));
printf("----------------------\n");
printf("Pre-order traversal\n");
post_traversal(root);
printf("----------------------\n");
printf("Post-order traversal\n");
pre_traversal(root);
printf("----------------------\n");
printf("Height of the tree is: %d\n", height(root));
printf("Least value in tree is: %d\n", least_of_tree(root));
printf("Max value of tree is: %d\n", max_of_tree(root));
int height_ = height(root);
int vertical_sum[10000];
for(int i = 0; i < 10000; ++i)
{
vertical_sum[i] = 0;
}
/*
I assumed Vertical sum for root is 1... :P if there are more no of nodes on the
left subtree of the root..they we cant implement this using arrays in c..as negative indices is not possible
*/
verticalSumArray(root,vertical_sum,1);
for(int i = 0;vertical_sum[i] != 0 ; ++i)
{
printf("%d : %d \n",i,vertical_sum[i]);
}
// tree_t mirrored_root = mirror_tree(root);
printf("----------------------\n");
// display(mirrored_root);
//printf("%d\n", RLMirrors(root));
//inorder(root);
tree_t largest = largestSumSubtree(root);
printf("%d\n",largest -> key);
int key = 1;
if(search(root,key) != NULL)
{
printf("%d is present\n",key);
}
else
{
printf("%d is absent\n",key);
}
iterativeInorder(root);
printf("----------------------\n");
iterativePreorder(root);
printf("----------------------\n");
iterativePostorder(root);
levelorder(root);
int height = 0;
printf("the diameter of the tree: %d\n",diameter(root));
int key = 9;
printAncestorsI(root,key);
int k = 3;
printKdistance(root,k);
int sum = 36;
if(hasSumPath(root,sum) == 1)
{
printf("Yes\n");
}
else
{
printf("No\n");
}
root = getNode(10);
root -> llink = getNode(8);
root -> rlink = getNode(2);
root -> llink -> llink = getNode(3);
root -> llink -> rlink = getNode(5);
root -> rlink ->rlink = getNode(2);
display(root);
printf("----------------------\n");
if(isSum(root))
{
printf("Yes\n");
}
else
{
printf("No\n");
}
root = getNode(20);
root -> llink = getNode(8);
root -> rlink = getNode(22);
root -> llink -> llink = getNode(4);
root -> llink -> rlink = getNode(12);
root -> llink -> rlink -> rlink = getNode(14);
root -> llink -> rlink -> llink = getNode(10);
int n1 = 4;
int n2 = 14;
printf("Least common ancestor of %d & %d is %d \n", n1,n2,lowestCommonAncestor(root,n1,n2));
printf("\n");
#endif
// printPaths(root);
#if 0
int num = 4;
printf("The no of bst for %d keys is %d\n", num,countTrees(num));i
#endif
#if 0
root = getNode(20);
root -> llink = getNode(8);
root -> rlink = getNode(22);
root -> llink -> llink = getNode(4);
root -> llink -> rlink = getNode(12);
root -> llink -> rlink -> llink = getNode(10);
root -> llink -> rlink -> rlink = getNode(14);
tree_t successor = inorderSuccessor(root,root -> llink);
if(successor != NULL)
{
printf("thei inorder successor is %d\n",successor -> key);
}
#endif
#if 0
root = getNode(20);
root -> llink = getNode(8);
root -> rlink = getNode(22);
root -> llink -> llink = getNode(100);
if(checkAllLeavesSameLevel(root) == 1)
{
printf("All the leaves are in the same level\n");
}
#endif
#if 0
display(root);
printf("-------------------------------------\n");
replaceNodeWithSum(root,0);
display(root);
int getSum(tree_t root)
{
if(root == NULL)
{
return 0;
}
else
{
return root -> key + getSum(root -> rlink);
}
}
#endif
// printf("%d\n",countTrees(3));
#if 0
root = getNode(10);
root->llink = getNode(20);
root->rlink = getNode(30);
if(isBalanced(root) == 1)
{
printf("The tree is balanced\n");
}
else
{
printf("Ther tree is unbalanced\n");
}
#endif
#if 0
tree_t root = NULL;
int array[] = {1,2,3,4,5,6,7,8,9};
int n = 9;
root = createMinimalBST(root,array,n);
display(root);
#endif
// tree_t root = NULL;
#if 0
root = getNode(20);
root -> llink = getNode(8);
root -> rlink = getNode(22);
root -> llink -> llink = getNode(4);
root -> llink -> rlink = getNode(12);
root -> llink -> rlink -> llink = getNode(10);
root -> llink -> rlink -> rlink = getNode(14);
// printPathWithSum(root,40);
// levelorderRecursive(root);
// rightView(root);
#endif
#if 0
int inorder[] = {4, 2, 5, 1, 6, 3};
int preorder[] = {1, 2, 4, 5, 3, 6};
int n = 6;
// int preIndex = 0;
// root = buildTreeWithInorderAndPreOrder(inorder,preorder,0,n - 1,&preIndex);
// display(root);
int index = 0;
printPostOrder(inorder,preorder,0,n - 1,&index);
#endif
#if 0
/*
10
-2 7
8 -4
*/
tree_t root = NULL;
root = getNode(10);
root->llink = getNode(-2);
root->rlink = getNode(7);
root->llink->llink = getNode(8);
root->llink->rlink = getNode(-4);
// printf("Sum: %d\n",maxSumPath(root));
pre_traversal(root);
printf("----------------------------------");
display(root);
#endif
return 0;
}
void testCountTrees() {
printf("Number of unique trees with values from 1 to 5 are %d \n",countTrees(5));
}