5.6 Implementing Tree Traversals
Recall that any recursive function requires the following:
- The base case and its action.
- The recursive case and its action.
In this module, we will talk about some details related to correctly and clearly implementing recursive tree traversals.
5.6.1 Base Case
In binary tree traversals, most often the base case is to check if we have an empty tree. A common mistake is to check the child pointers of the current node, and only make the recursive call for a non-null child.
Recall the basic preorder traversal function.
function preorder(node):
if node is not null: // Only continue if this is a tree
// Visit root node
visit(node) // Process all nodes in left subtree
preorder(node.left) // Process all nodes in right subtree preorder(node.right)
Here is an alternate design for the preorder traversal, in which the left and right pointers of the current node are checked so that the recursive call is made only on non-empty children.
// This is a bad idea:
function preorder2(node):
visit(node)if node.left is not null:
preorder2(node.left)if node.right is not null:
preorder2(node.right)
At first, it might appear that preorder2
is more
efficient than preorder
, because it makes only half as many
recursive calls (since it won’t try to call on a null pointer). On the
other hand, preorder2
must access the left and right child
pointers twice as often. The net result is that there is no performance
improvement.
Perhaps the writer of preorder2
wants to protect against
the case where the root is null
. But preorder2
has an error. While preorder2
insures that no recursive
calls will be made on empty subtrees, it will fail if the original call
from outside passes in a null pointer. This would occur if the original
tree is empty. Since an empty tree is a legitimate input to the initial
call on the function, there is no safe way to avoid this case. So it is
necessary that the first thing you do on a binary tree traversal is to
check that the root is not null
. If we try to fix
preorder2
by adding this test, then making the tests on the
children is completely redundant because the pointer will be checked
again in the recursive call.
The design of preorder2
is inferior to that of
preorder
for a deeper reason as well. Looking at the
children to see if they are null
means that we are worrying
too much about something that can be dealt with just as well by the
children. This makes the function more complex, which can become a real
problem for more complex tree structures. Even in the relatively simple
preorder2
function, we had to write two tests for
null
rather than the one needed by preorder
.
This makes it more complicated than the original version. The key issue
is that it is much easier to write a recursive function on a tree when
we only think about the needs of the current node. Whenever we can, we
want to let the children take care of themselves. In this case, we care
that the current node is not null
, and we care about how to
invoke the recursion on the children, but we do not
have to care about how or when that is done.
5.6.2 The Recursive Call
The secret to success when writing a recursive function is to not worry about how the recursive call works. Just accept that it will work correctly. One aspect of this principle is not to worry about checking your children when you don’t need to. You should only look at the values of your children if you need to know those values in order to compute some property of the current node. Child values should not be used to decide whether to call them recursively. Make the call, and let their own base case handle it.
Example: Changing the node values in a tree
Consider the problem of incrementing the value for each node in a binary tree. The following solution has an error, since it does redundant manipulation to the left and the right children of each node.
function inefficient_increment(node):
if node is not null:
= node.elem + 1
node.elem if node.left is not null:
= node.left.elem + 1
node.left.elem
inefficient_increment(node.left.left)if node.right is not null:
= node.right.elem + 1
node.right.elem inefficient_increment(node.right.right)
The efficient solution should not explicitly set the child values that way. Changing the value of a node does not depend on the child values. So the function should simply increment the node value, and make recursive calls on the children.
In rare problems, you might need to explicitly check if the children are null or access the child values for each node. For example, you might need to check if all nodes in a tree satisfy the property that each node stores the sum of its left and right children. In this situation, you must look at the values of the children to decide something about the current node. You do not look at the children to decide whether to make a recursive call.