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
Abigail
Angela
Adela
Adam
Alice
Agnes
Audrey
Allen
Arthur
Binary Search Tree (BST): Implementation
struct 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 version
int 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
Abigail
Angela
Adela
Adam
Alice
Agnes
Audrey
Allen
Arthur
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 Node
struct 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
Abigail
Angela
Adela
Adam
Alice
Agnes
Audrey
Allen
Arthur
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 tree
Node 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 code
struct 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 lmc
EleType leftMostChild (struct node * current) {
while (current->left != 0)
current = current->left;
return current->value;
}
Second helper routine - remove leftmost
struct 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??