7.13. Search Tree Implementation¶
A binary search tree relies on the property that keys that are less than the parent are found in the left subtree, and keys that are greater than the parent are found in the right subtree. We will call this the bst property. As we implement the Map interface as described above, the bst property will guide our implementation. Figure 1 illustrates this property of a binary search tree, showing the keys without any associated values. Notice that the property holds for each parent and child. All of the keys in the left subtree are less than the key in the root. All of the keys in the right subtree are greater than the root.
Now that you know what a binary search tree is, we will look at how a binary search tree is constructed. The search tree in Figure 1 represents the nodes that exist after we have inserted the following keys in the order shown: \(70,31,93,94,14,23,73\). Since 70 was the first key inserted into the tree, it is the root. Next, 31 is less than 70, so it becomes the left child of 70. Next, 93 is greater than 70, so it becomes the right child of 70. Now we have two levels of the tree filled, so the next key is going to be the left or right child of either 31 or 93. Since 94 is greater than 70 and 93, it becomes the right child of 93. Similarly 14 is less than 70 and 31, so it becomes the left child of 31. 23 is also less than 31, so it must be in the left subtree of 31. However, it is greater than 14, so it becomes the right child of 14.
To implement the binary search tree, we will use the nodes and
references approach similar to the one we used to implement the linked
list, and the expression tree. However, because we must be able create
and work with a binary search tree that is empty, our implementation
will use two classes. The first class we will call
and the second class we will call
class has a reference to the
TreeNode that is the root of the binary
search tree. In most cases the external methods defined in the outer
class simply check to see if the tree is empty. If there are nodes in
the tree, the request is just passed on to a private method defined in
BinarySearchTree class that takes the root as a parameter. In
the case where the tree is empty or we want to delete the key at the
root of the tree, we must take special action. The code for the
BinarySearchTree class constructor along with a few other
miscellaneous functions is shown in Listing 1.
class BinarySearchTree: def __init__(self): self.root = None self.size = 0 def length(self): return self.size def __len__(self): return self.size def __iter__(self): return self.root.__iter__()
TreeNode class provides many helper functions that make the work
done in the
BinarySearchTree class methods much easier. The
constructor for a
TreeNode, along with these helper functions, is
shown in Listing 2. As you can see in the listing many of
these helper functions help to classify a node according to its own
position as a child, (left or right) and the kind of children the node
TreeNode class will also explicitly keep track
of the parent as an attribute of each node. You will see why this is
important when we discuss the implementation for the
Another interesting aspect of the implementation of
Listing 2 is that we use Python’s optional parameters.
Optional parameters make it easy for us to create a
several different circumstances. Sometimes we will want to construct a
TreeNode that already has both a
parent and a
With an existing parent and child, we can pass parent and child as
parameters. At other times we will just create a
TreeNode with the
key value pair, and we will not pass any parameters for
child. In this case, the default values of the optional parameters
class TreeNode: def __init__(self,key,val,left=None,right=None, parent=None): self.key = key self.payload = val self.leftChild = left self.rightChild = right self.parent = parent def hasLeftChild(self): return self.leftChild def hasRightChild(self): return self.rightChild def isLeftChild(self): return self.parent and self.parent.leftChild == self def isRightChild(self): return self.parent and self.parent.rightChild == self def isRoot(self): return not self.parent def isLeaf(self): return not (self.rightChild or self.leftChild) def hasAnyChildren(self): return self.rightChild or self.leftChild def hasBothChildren(self): return self.rightChild and self.leftChild def replaceNodeData(self,key,value,lc,rc): self.key = key self.payload = value self.leftChild = lc self.rightChild = rc if self.hasLeftChild(): self.leftChild.parent = self if self.hasRightChild(): self.rightChild.parent = self
Now that we have the
BinarySearchTree shell and the
is time to write the
put method that will allow us to build our
binary search tree. The
put method is a method of the
BinarySearchTree class. This method will check to see if the tree
already has a root. If there is not a root then
put will create a
TreeNode and install it as the root of the tree. If a root node
is already in place then
put calls the private, recursive, helper
_put to search the tree according to the following
Starting at the root of the tree, search the binary tree comparing the new key to the key in the current node. If the new key is less than the current node, search the left subtree. If the new key is greater than the current node, search the right subtree.
When there is no left (or right) child to search, we have found the position in the tree where the new node should be installed.
To add a node to the tree, create a new
TreeNodeobject and insert the object at the point discovered in the previous step.
Listing 3 shows the Python code for inserting a new node in
the tree. The
_put function is written recursively following the
steps outlined above. Notice that when a new child is inserted into the
currentNode is passed to the new tree as the parent.
One important problem with our implementation of insert is that duplicate keys are not handled properly. As our tree is implemented a duplicate key will create a new node with the same key value in the right subtree of the node having the original key. The result of this is that the node with the new key will never be found during a search. A better way to handle the insertion of a duplicate key is for the value associated with the new key to replace the old value. We leave fixing this bug as an exercise for you.
def put(self,key,val): if self.root: self._put(key,val,self.root) else: self.root = TreeNode(key,val) self.size = self.size + 1 def _put(self,key,val,currentNode): if key < currentNode.key: if currentNode.hasLeftChild(): self._put(key,val,currentNode.leftChild) else: currentNode.leftChild = TreeNode(key,val,parent=currentNode) else: if currentNode.hasRightChild(): self._put(key,val,currentNode.rightChild) else: currentNode.rightChild = TreeNode(key,val,parent=currentNode)
put method defined, we can easily overload the
operator for assignment by having the
__setitem__ method call (see Listing 4) the
put method. This allows us to write Python statements like
myZipTree['Plymouth'] = 55446, just like a Python dictionary.
def __setitem__(self,k,v): self.put(k,v)
Figure 2 illustrates the process for inserting a new node into a binary search tree. The lightly shaded nodes indicate the nodes that were visited during the insertion process.
Once the tree is constructed, the next task is to implement the
retrieval of a value for a given key. The
get method is even easier
put method because it simply searches the tree recursively
until it gets to a non-matching leaf node or finds a matching key. When
a matching key is found, the value stored in the payload of the node is
Listing 5 shows the code for
__getitem__. The search code in the
_get method uses the same
logic for choosing the left or right child as the
_put method. Notice
_get method returns a
get, this allows
_get to be used as a flexible helper method for other
BinarySearchTree methods that may need to make use of other data
TreeNode besides the payload.
By implementing the
__getitem__ method we can write a Python
statement that looks just like we are accessing a dictionary, when in
fact we are using a binary search tree, for example
z = myZipTree['Fargo']. As you can see, all the
__getitem__ method does is call
def get(self,key): if self.root: res = self._get(key,self.root) if res: return res.payload else: return None else: return None def _get(self,key,currentNode): if not currentNode: return None elif currentNode.key == key: return currentNode elif key < currentNode.key: return self._get(key,currentNode.leftChild) else: return self._get(key,currentNode.rightChild) def __getitem__(self,key): return self.get(key)
get, we can implement the
in operation by writing a
__contains__ method for the
__contains__ method will simply call
get and return
get returns a value, or
False if it returns
__contains__ is shown in Listing 6.
def __contains__(self,key): if self._get(key,self.root): return True else: return False
__contains__ overloads the
in operator and allows us
to write statements such as:
if 'Northfield' in myZipTree: print("oom ya ya")
Finally, we turn our attention to the most challenging method in the
binary search tree, the deletion of a key (see Listing 7). The first task is to find the
node to delete by searching the tree. If the tree has more than one node
we search using the
_get method to find the
TreeNode that needs
to be removed. If the tree only has a single node, that means we are
removing the root of the tree, but we still must check to make sure the
key of the root matches the key that is to be deleted. In either case if
the key is not found the
del operator raises an error.
def delete(self,key): if self.size > 1: nodeToRemove = self._get(key,self.root) if nodeToRemove: self.remove(nodeToRemove) self.size = self.size-1 else: raise KeyError('Error, key not in tree') elif self.size == 1 and self.root.key == key: self.root = None self.size = self.size - 1 else: raise KeyError('Error, key not in tree') def __delitem__(self,key): self.delete(key)
Once we’ve found the node containing the key we want to delete, there are three cases that we must consider:
The node to be deleted has no children (see Figure 3).
The node to be deleted has only one child (see Figure 4).
The node to be deleted has two children (see Figure 5).
The first case is straightforward (see Listing 8). If the current node has no children all we need to do is delete the node and remove the reference to this node in the parent. The code for this case is shown in here.
if currentNode.isLeaf(): if currentNode == currentNode.parent.leftChild: currentNode.parent.leftChild = None else: currentNode.parent.rightChild = None
The second case is only slightly more complicated (see Listing 9). If a node has only a single child, then we can simply promote the child to take the place of its parent. The code for this case is shown in the next listing. As you look at this code you will see that there are six cases to consider. Since the cases are symmetric with respect to either having a left or right child we will just discuss the case where the current node has a left child. The decision proceeds as follows:
If the current node is a left child then we only need to update the parent reference of the left child to point to the parent of the current node, and then update the left child reference of the parent to point to the current node’s left child.
If the current node is a right child then we only need to update the parent reference of the left child to point to the parent of the current node, and then update the right child reference of the parent to point to the current node’s left child.
If the current node has no parent, it must be the root. In this case we will just replace the
rightChilddata by calling the
replaceNodeDatamethod on the root.
else: # this node has one child if currentNode.hasLeftChild(): if currentNode.isLeftChild(): currentNode.leftChild.parent = currentNode.parent currentNode.parent.leftChild = currentNode.leftChild elif currentNode.isRightChild(): currentNode.leftChild.parent = currentNode.parent currentNode.parent.rightChild = currentNode.leftChild else: currentNode.replaceNodeData(currentNode.leftChild.key, currentNode.leftChild.payload, currentNode.leftChild.leftChild, currentNode.leftChild.rightChild) else: if currentNode.isLeftChild(): currentNode.rightChild.parent = currentNode.parent currentNode.parent.leftChild = currentNode.rightChild elif currentNode.isRightChild(): currentNode.rightChild.parent = currentNode.parent currentNode.parent.rightChild = currentNode.rightChild else: currentNode.replaceNodeData(currentNode.rightChild.key, currentNode.rightChild.payload, currentNode.rightChild.leftChild, currentNode.rightChild.rightChild)
The third case is the most difficult case to handle (see Listing 10). If a node has two children, then it is unlikely that we can simply promote one of them to take the node’s place. We can, however, search the tree for a node that can be used to replace the one scheduled for deletion. What we need is a node that will preserve the binary search tree relationships for both of the existing left and right subtrees. The node that will do this is the node that has the next-largest key in the tree. We call this node the successor, and we will look at a way to find the successor shortly. The successor is guaranteed to have no more than one child, so we know how to remove it using the two cases for deletion that we have already implemented. Once the successor has been removed, we simply put it in the tree in place of the node to be deleted.
The code to handle the third case is shown in the next listing.
Notice that we make use of the helper methods
findMin to find the successor. To remove the successor, we make use
of the method
spliceOut. The reason we use
spliceOut is that it
goes directly to the node we want to splice out and makes the right
changes. We could call
delete recursively, but then we would waste
time re-searching for the key node.
elif currentNode.hasBothChildren(): #interior succ = currentNode.findSuccessor() succ.spliceOut() currentNode.key = succ.key currentNode.payload = succ.payload
The code to find the successor is shown below (see Listing 11) and as
you can see is a method of the
TreeNode class. This code makes use
of the same properties of binary search trees that cause an inorder
traversal to print out the nodes in the tree from smallest to largest.
There are three cases to consider when looking for the successor:
If the node has a right child, then the successor is the smallest key in the right subtree.
If the node has no right child and is the left child of its parent, then the parent is the successor.
If the node is the right child of its parent, and itself has no right child, then the successor to this node is the successor of its parent, excluding this node.
The first condition is the only one that matters for us when deleting a
node from a binary search tree. However, the
has other uses that we will explore in the exercises at the end of this
findMin method is called to find the minimum key in a subtree.
You should convince yourself that the minimum valued key in any binary
search tree is the leftmost child of the tree. Therefore the
method simply follows the
leftChild references in each node of the
subtree until it reaches a node that does not have a left child.
def findSuccessor(self): succ = None if self.hasRightChild(): succ = self.rightChild.findMin() else: if self.parent: if self.isLeftChild(): succ = self.parent else: self.parent.rightChild = None succ = self.parent.findSuccessor() self.parent.rightChild = self return succ def findMin(self): current = self while current.hasLeftChild(): current = current.leftChild return current def spliceOut(self): if self.isLeaf(): if self.isLeftChild(): self.parent.leftChild = None else: self.parent.rightChild = None elif self.hasAnyChildren(): if self.hasLeftChild(): if self.isLeftChild(): self.parent.leftChild = self.leftChild else: self.parent.rightChild = self.leftChild self.leftChild.parent = self.parent else: if self.isLeftChild(): self.parent.leftChild = self.rightChild else: self.parent.rightChild = self.rightChild self.rightChild.parent = self.parent
We need to look at one last interface method for the binary search tree.
Suppose that we would like to simply iterate over all the keys in the
tree in order. This is definitely something we have done with
dictionaries, so why not trees? You already know how to traverse a
binary tree in order, using the
inorder traversal algorithm.
However, writing an iterator requires a bit more work, since an iterator
should return only one node each time the iterator is called.
Python provides us with a very powerful function to use when creating an
iterator. The function is called
yield is similar to
return in that it returns a value to the caller. However,
also takes the additional step of freezing the state of the function so
that the next time the function is called it continues executing from
the exact point it left off earlier. Functions that create objects that
can be iterated are called generator functions.
The code for an
inorder iterator of a binary tree is shown in the next
listing. Look at this code carefully; at first glance you
might think that the code is not recursive. However, remember that
__iter__ overrides the
for x in operation for iteration, so it
really is recursive! Because it is recursive over
__iter__ method is defined in the
def __iter__(self): if self: if self.hasLeftChild(): for elem in self.leftChiLd: yield elem yield self.key if self.hasRightChild(): for elem in self.rightChild: yield elem
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