CS 261 – Fall 2009

Binary Search Trees

Again, but in detail

Binary Search Tree•Binary search trees are binary tree’s where every node’s object value is:–Greater than or equal to all its descendents in the left subtree

–Less than or equal to all its descendents in the right subtree

•An in-order traversal will return the elements in sorted order

•If the tree is reasonably full, searching for an element is O(log n)

Binary Search Tree: Example

Alex

Abner Angela

Adela Alice Audrey

Adam Agnes Allen Arthur

Abigail

Binary Search Tree (BST): Implementationstruct BST { struct Node * root; int size;};

struct node { EleType value;

struct node * left;

struct node * right;

};

void BSTinit (struct BST *tree);

void BSTadd(struct BST *tree, EleType value);

int BSTcontains (struct BST *tree, EleType value);

void BSTremove (struct BST *tree, EleType value);

int BSTsze (struct BST *tree);

Init - what needs to be done?struct BST { struct node * root; int size;};

void BSTinit (struct BST *tree) {

tree->root = 0;

tree->size = 0;

}

Implementing the Bag: Contains•Start at the root.

•At each node, compare to test: return true if match

•If test is less than node, look at left child

•Otherwise if test is greater than node, look at right child

•Traverses a path from the root to the leaf.

•Therefore, if the tree is reasonably complete (an important if) the execution time is O( ?? )

Use Recursion, or use a loop?•Both will work. Lets compare the two

int BSTcontains

(struct BST *tree, EleType value) {

struct node * current = tree->root;

while (current != 0) {

if (EQ(value, current->value)) return 1;

if (LT(value, current->value))

current = current->left;

else current = current->right;

}

return 0;

}

Recursive versionint BSTcontains

(struct BST *tree, EleType value)

{ return BSTnodecontains(tree->root, value); }

int BSTnodeContains

(struct BST *current, EleType value) {

if (current == 0) return 0;

if (EQ(value, current->value)) return 1;

if (LT(value, current->value))

return BSTnodeContains(current->left, value);

else return BSTnodeContains(current->right, value);

}

Which is easier to understand?•Somewhat a matter of style

•It is what you are used to

•Execution time will be very similar

Implementing a Bag: Add•Do the same type of traversal from root to leaf.

•When you find a null value, create a new node.

Alex

Abner Angela

Adela Alice Audrey

Adam Agnes Allen Arthur

Abigail

A useful trick•A useful trick (adapted from the functional programming world). Make a secondary routine that returns the tree with the value inserted.

Node add (Node current, EleType newValue) if current is null then return new Node with value

otherwise if newValue < Node value left child = add (left child, newValue) else right child = add (right child, newValue)

return current node

Add just calls the utility routine

void BSTadd (EleType newValue) {

root = _BSTnodeAdd(root, newValue);

dataSize++;

}

Real work is done inside Nodestruct node * BSTnodeAdd (struct node * current, EleType value) {

if (current == 0) {

current = (struct node *) malloc(sizeof(struct node));

assert(current != 0);

current->value = value;

current->left = current->right = 0;

} else if (LT(value, current->value))

current->left = BSTnodeAdd(current->left, value);

else

current->right = BSTnodeAdd(current->right, value);

return current;

}

Notes on Add•See how it returns the Current value, the tree AFTER the insertion is made

•Again, could be done with a loop - some people find the recursive version easier, some people find the loop easier.

•Try writing it using a loop

Bag Implementation: Remove•As is often the case, remove is the most complex. Leaves a “hole”. What value should fill the hole?

Alex

Abner Angela

Adela Alice Audrey

Adam Agnes Allen Arthur

Abigail

Who can fill that hole?•Answer: The Leftmost child of the right child. (Smallest element in right subtree)

•Try this on a few values.

•Useful to have a couple of private inner routines

EleType leftmostChild (Node current) {

… // return value of leftmost child of current

}

Struct Node * removeLeftmostChild (Node current) {

… // return tree with leftmost child removed

}

One additional special case•We have said you want to fill hole with leftmost child of right child. What if you don’t have a right child? Think about it. Can just return left child. Try remove “Audrey”Angela

Alice Audrey

Allen Arthur

Once more, separate root and node

void BSTremove (struct BST * tree, EleType value) {

if (BSTincludes(tree, value)) {

tree->root = BSTnodeRemove (tree->root, value);

tree->size--;

}

}

Node Remove: easy if you return a treeNode remove (Node current, EleType testValue)

if current->value is the thing we seek

if right child is null return left child

else replace value with leftmost child of right child

and set right child to be removeLeftmost (right)

else if testValue < current.value

left child = remove (left child, testValue)

else

right child = remove (right child, testValue)

return current node

Translate the previous into codestruct BSTnodeRemove (struct node * current, EleType testValue) {

if (EQ(testValue, current->value)) { /* found it! */

if (current->right == 0) return current->left;

else {

current->value = leftmostChild(current->right);

current->right = removeLeftMost (current->right); }

} else if (LT(testValue, current.value))

current->left = BSTnodeRemove (current->left, testValue)

else

current->right = BSTnoderemove (current->right, testValue)

return current;

}

Little Helper Routines - get value of lmcEleType leftMostChild (struct node * current) {

while (current->left != 0)

current = current->left;

return current->value;

}

Second helper routine - remove leftmoststruct node * removeLeftMost (struct node * current) {

if (current->left == 0)

return current->right;

else

current->left = removeLeftMost (current->left);

return current;

}

What is the complexity??Basically once more just running down a path from root to leaf

What is the complexity? O(???)

Careful! What can go wrong? What happens in the worst case??