Question

The following procedure list->tree converts an ordered list to a balanced binary tree. The helper procedure partial-tree takes as arguments an integer \(n\) and list of at least \(n\) elements and constructs a balanced tree containing the first \(n\) elements of the list. The result returned by partial- tree is a pair (formed with cons) whose car is the constructed tree and whose cdr is the list of elements not included in the tree. a. Write a short paragraph explaining as clearly as you can how partial-tree works. Draw the tree produced by list->tree for the list ( \(\left.\begin{array}{llllll}1 & 3 & 5 & 7 & 9 & 11\end{array}\right)\). b. What is the order of growth in the number of steps required by list->tree to convert a list of \(n\) elements?

Short answer

list->tree constructs a balanced tree with a time complexity of \(O(n \log n)\).

Step-by-step solution

Understanding partial-tree Function

The function `partial-tree` operates recursively to construct a balanced binary tree from the first `n` elements of a list. It splits the task into two sub-problems of equal size (where possible), corresponding to the left and right subtrees. The middle element of the list becomes the root of the tree. The function returns the constructed tree and the remainder of the list not included in the tree.

Analyzing the Example List

To draw the tree produced by `list->tree` for the list \([1, 3, 5, 7, 9, 11]\), we need to split the list into sublists corresponding to the left subtree, the root, and the right subtree. The middle element (5) becomes the root of the tree. The left subtree is built from \([1, 3]\) and the right subtree from \([7, 9, 11]\).

Building the Left Subtree

The left subtree with elements \([1, 3]\) places 3 as the root (middle element), with 1 as its left child (as there's no right side at this point).

Building the Right Subtree

The right subtree with elements \([7, 9, 11]\) has 9 as the root (since it is the middle element), with 7 as the left child and 11 as the right child.

Combining Subtrees into the Complete Tree

The complete tree constructed from the list \([1, 3, 5, 7, 9, 11]\) is balanced with 5 as the root, creating a subtree structure as follows: 5 / \ 3 9 / / \ 1 7 11

Determining Order of Growth

The order of growth for `list->tree` arises from dividing the list approximately in half at each level of the recursion, similar to a binary search structure. Each division reduces the problem size, leading to an overall time complexity of \(O(n \log n)\) due to visiting each element of the list during merge stages and tree construction process.

Key concepts

Recursion

Recursion is a fundamental programming concept where a function calls itself to solve smaller instances of a problem. In the context of converting a list to a balanced binary tree, recursion is employed to handle the task of constructing the tree efficiently.
The `partial-tree` function is recursive and divides its input list into two halves—a left subtree and a right subtree—during each recursive invocation. By finding the middle element, it assigns this element as the root of the subtree, allowing the left and right sub-lists to become smaller subproblems.
Each recursive call thus processes part of the list, ultimately constructing the binary tree in a balanced manner. The benefit of recursion here lies in its neat organization: each part of the tree is constructed as soon as the function processes its corresponding sub-list. Eventually, the recursive calls culminate in returning a balanced binary tree structured based on the initial input list.

Order of Growth

Understanding the order of growth in algorithms helps us estimate their efficiency and resource usage as the input size increases. For the `list->tree` procedure, the order of growth is significant.
The list is divided approximately in half with each recursive step, somewhat akin to a binary search mechanic. As a result, the height of the tree becomes logarithmically related to the input size, hence impacting the time complexity.
Since each element of the list needs to be visited and used in constructing the tree, combining this constant factor with the divide-and-conquer nature of the algorithm results in a time complexity of \(O(n \log n)\). The \(n\) factor accounts for each element's visitation, while the \(\log n\) factor arises from the recursive division of the input list.

Tree Construction

Tree construction in this context involves creating a balanced binary tree from a given ordered list. The process utilizes recursion to maintain balance by ensuring the depths of two child subtrees of any node differ by at most one.
The key part of tree construction lies in selecting the middle element of the list as the root of the subtree. This choice helps maintain the balance criterion. With each recursive step constructing left and right subtrees, the function effectively maintains a systematic and balanced structure.
For example, if you have a list like \([1, 3, 5, 7, 9, 11]\), the algorithm will choose 5 as the root. The left subtree will include [1, 3] and the right [7, 9, 11], thus maintaining balance. This recursive breakdown ensures that each node fittingly represents an ordered structure of the original list.

Binary Search

Binary search is an efficient algorithm for finding an item from a sorted list. It significantly resembles the methodology of building a balanced binary tree, as both involve dividing the data into halves for efficient processing.
When list elements are sorted and a binary search needs to be performed, it can operate on a binary tree wherein each step involves comparing the middle element of the subtree to the target value, just like choosing the middle element during tree construction to be the root of the subtree.
In constructing a balanced binary tree, this middle-element selection ensures the tree is depth balanced, directly mirroring the binary search strategy. At each step, just like binary search narrows down the range, selecting the middle ensures even distribution across both sides of the tree, optimizing insertion and search operations for future data retrieval from the tree.