Avl Tree Practice Questions

To determine the worst-case time complexity, consider these points:

1. An AVL tree with $n$ elements has a height of $\Theta(\log n)$.
2. Inserting a single element into an AVL tree takes $\Theta(\log k)$ time, where $k$ is the current number of elements in the tree. This involves finding the insertion point and potentially performing rotations.
3. We are inserting $n^2$ new elements into an AVL tree that initially has $n$ elements.
4. The total time for inserting $n^2$ elements will be the sum of the time taken for each individual insertion. As the tree grows from $n$ to $n+n^2$ elements, the cost of each insertion will be at most $\Theta(\log (n+n^2)) = \Theta(\log (n^2)) = \Theta(2 \log n) = \Theta(\log n)$.
5. Since we perform $n^2$ insertions, each taking $\Theta(\log n)$ time in the worst case (when the tree is largest), the total worst-case time complexity is $n^2 \times \Theta(\log n) = \Theta(n^2 \log n)$.

To report all elements in a balanced binary search tree within a given range $[a, b]$, the process involves two main steps:

1. Finding the range boundaries: Locate the elements $a$ and $b$ (or their closest existing values) in the BST. Since the tree is balanced, searching for these two boundary elements takes $\Theta(\log n)$ time.
2. Reporting elements within the range: Once $a$ and $b$ are found, an in-order traversal can be performed starting from $a$ and continuing until $b$ is reached. Each element reported during this traversal takes constant time. If $k$ elements are reported, this step takes $\Theta(k)$ time.

Combining these steps, the total worst-case time complexity is $\Theta(\log n + k)$.

To achieve logarithmic access time for a symbol table, the data structure must allow for the repeated division of the search space. In a balanced binary search tree (a type of search tree), each comparison effectively narrows the search area by approximately half. For $n$ total elements, the height of the tree is $h \approx \log_2(n)$. Consequently, the time complexity for searching within such a tree is $O(\log n)$. In contrast, linear lists typically require $O(n)$ time, hash tables target $O(1)$ average time, and self-organizing lists generally result in $O(n)$ worst-case time. Thus, the search tree provides the required logarithmic performance.

To find the sum of numbers between L and H in a balanced binary search tree:

1. Find the smallest number greater than or equal to L. This takes $O(\log n)$ time.
2. Find the largest number less than or equal to H. This also takes $O(\log n)$ time.
3. Traverse the tree between these two numbers to sum the $m$ elements. This takes $O(m)$ time.
   Thus, the total time complexity is $O(\log n + m)$.
   Comparing this with $O(n^{a}\log^{b}n+m^{c}\log^{d}n)$, we get:
   $a=0$, $b=1$ (from $\log n$)
   $c=1$, $d=0$ (from $m$)
   The value of $a + 10b + 100c + 1000d = 0 + 10(1) + 100(1) + 1000(0) = 10 + 100 = 110$.

[CORRECT_OPTION: 110]

To determine the number of rotations, we insert elements into the AVL tree step-by-step:

1. Insert 9, 6, 5: Insertion of $5$ creates a Left-Left ($LL$) imbalance at $9$. A single right rotation at $9$ restores balance. Tree: $6$ (root), $5$ (left), $9$ (right). (1 rotation)
2. Insert 8: Added as the left child of $9$. Tree: $6$ (root), $5$ (left), $9$ (right), $8$ (left of $9$).
3. Insert 7: Added as the left child of $8$. This creates a $LL$ imbalance at $9$. A single right rotation at $9$ restores balance. Tree: $6$ (root), $5$ (left), $8$ (right), $7$ (left of $8$), $9$ (right of $8$). (1 rotation)
4. Insert 10: Added as the right child of $9$. Now, $BF(6) = -2$ (a Right-Right imbalance). A single left rotation at $6$ balances the tree. (1 rotation)

Total rotations = $1 + 1 + 1 = 3$.

In a balanced binary search tree, the worst-case running time to search for an element is proportional to its height.
Given that the number of elements in the tree is $N = n2^n$.
The height of a balanced binary search tree with $N$ elements is $O(\log N)$.
Substituting $N = n2^n$ into the height formula, we get:
$O(\log(n2^n)) = O(\log n + \log(2^n)) = O(\log n + n \log 2) = O(\log n + n)$
Since $n$ grows faster than $\log n$, the dominant term is $n$.
Therefore, the worst-case running time is $\Theta(n)$.

Let $N_h$ be the minimum number of nodes in an AVL tree of height $h$.
Given that the height of a single node is 0:
$N_0 = 1$ (a single node)
$N_1 = 2$ (a root with one child)

The recurrence relation for the minimum number of nodes in an AVL tree of height $h$ is $N_h = N_{h-1} + N_{h-2} + 1$.
Using this, we calculate:
$N_0 = 1$
$N_1 = 2$
$N_2 = N_1 + N_0 + 1 = 2 + 1 + 1 = 4$
$N_3 = N_2 + N_1 + 1 = 4 + 2 + 1 = 7$
$N_4 = N_3 + N_2 + 1 = 7 + 4 + 1 = 12$
An AVL tree with $N$ nodes can have a maximum height $h$ such that $N_h \le N < N_{h+1}$. For $N=7$, we find $N_3=7$ and $N_4=12$. Since $N_3 \le 7 < N_4$, the maximum height is 3.

For a balanced binary search tree with $n$ leaf nodes, the number of leaf nodes in the left subtree $n_L$ and right subtree $n_R$ are both approximately $n/2$. The recurrence relation for the total cost $T(n)$ is:
$T(n) = T(n_L) + T(n_R) + \min(n_L, n_R)$
Substituting $n_L = n_R = n/2$, we get:
$T(n) = 2T(n/2) + n/2$
Applying the Master Theorem, where $a=2$, $b=2$, and $f(n) = \Theta(n)$, we compare $f(n)$ with $n^{\log_b a} = n^{\log_2 2} = n^1$. Since $f(n) = \Theta(n^1)$, this falls under Case 2 of the Master Theorem.
Thus, the time complexity is $T(n) = \Theta(n \log n)$.

To maximize the height $h$ of a weight-balanced tree with $n$ nodes, we minimize the number of nodes $N(h)$ required to reach that height. For a tree of height $h$, let the number of nodes in the left subtree be $L$ and in the right subtree be $R$. To minimize $N(h)$, we set one subtree to the minimum size for height $h-1$ and the other to the minimum possible size that satisfies the balance constraint $L \le 2R$ (or $R \le 2L$).

By setting $L = N(h-1)$ and $R = \frac{1}{2} L = \frac{1}{2} N(h-1)$, the recurrence relation is:
$N(h) = N(h-1) + \frac{1}{2} N(h-1) + 1 = \frac{3}{2} N(h-1) + 1$
Ignoring the constant term for the asymptotic growth, we have $N(h) \approx \left(\frac{3}{2}\right)^h$. Taking the logarithm gives $h \approx \log_{3/2} n$.

In an AVL tree, the balance factor ($BF$) is defined as $\text{height}(\text{left}) - \text{height}(\text{right})$. A node is unbalanced if $|BF| \ge 2$.

After inserting a child node under $g$, the heights of ancestors change: node $g$ reaches height $1$, node $c$ reaches height $2$, node $b$ reaches height $3$, and node $a$ reaches height $4$. We calculate the new $BF$ for these ancestors:

• $BF(c) = \text{height}(g) - \text{height}(\text{null}) = 1 - (-1) = 2$ (Unbalanced)
• $BF(b) = \text{height}(c) - \text{height}(d) = 2 - 0 = 2$ (Unbalanced)
• $BF(a) = \text{height}(b) - \text{height}(e) = 3 - 1 = 2$ (Unbalanced)

Since $BF(c)$, $BF(b)$, and $BF(a)$ are all $2$, a total of $3$ nodes become unbalanced.