Say You Set The Key For Each Position Of A Binary Tree Equal

Say You Set The Key For Each Positionpof A Binary Treetequal To Its

Say you set the key for each position p of a binary tree T equal to its pre-order rank. Under what circumstances is T a heap? In other words, what property should T have to be a heap? Discuss about such a property. If we are to insert "43," "18," and "2," on the binary tree shown below (it's also a heap!), what is the end result? Remember we need three steps to complete inserting in a heap: a) Place the new element in the next available position in the array. b) Compare the new element with its parent. If the new element is smaller, than swap it with its parent. c) Continue this process until either the new element’s parent is smaller than the new element or it reaches the root.

Consider a red-black tree (RBT) T storing 1,024 elements. What is the worst-case height of T? As you recall, the height is the black height. You do not have to be correct in terms of the answer, but try to show the reason why it should be that number 300 words apa format 2 sources do not include the prompt in the final output.

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Setting the keys of a binary tree according to the pre-order rank of each node introduces specific structural properties. When such a binary tree T qualifies as a heap, it must satisfy the heap property: for every node, the key must be greater than or equal to the keys of its children in a max-heap, or less than or equal in a min-heap. Given that the keys are assigned based on pre-order traversal, the critical question is whether the traversal order aligns with the heap property. Specifically, if T is a max-heap, the pre-order rank must be arranged such that nodes are visited in decreasing order of their keys, ensuring parent nodes have higher keys than their children. This arrangement is feasible if and only if the parent’s pre-order rank is less than that of its children, which correlates with a specific order of node placement and potential heapifying conditions. In such a scenario, the pre-order rank directly influences whether T maintains heap properties, especially if the traversal sequence encapsulates the heap's max- or min-conditions.

Regarding insertion into a heap using the specified process: after placing the element at the next available position, comparisons and swaps ensure the heap property is maintained. For example, inserting "43," "18," and "2" step-by-step into the heap involves placing each element at the end of the heap array, then bubbling it up by comparing it with its parent and swapping if necessary. These steps are repeated until the parent is smaller or the root is reached. The final structure depends on the initial heap shape but will always satisfy heap properties after each insertion, ensuring the maximum or minimum element percolates appropriately.

For understanding the height of a red-black tree with 1,024 nodes, it is important to consider the balancing properties of such trees. A red-black tree guarantees that the longest path from the root to a leaf is no more than twice the length of the shortest path, thanks to strict balancing rules involving black nodes and red nodes. The black-height, defined as the number of black nodes from the root to a leaf, influences the overall height of the tree. For 1,024 elements, the worst-case black-height can be approximated by using logarithmic calculations. As a red-black tree maintains a black-height proportional to log2(n+1), the maximum black-height for 1,024 nodes would be around log2(1024), which equals 10, plus a factor accounting for red nodes, resulting in a worst-case height of approximately 20. This ensures efficient search, insertion, and deletion operations in logarithmic time, illustrating the effectiveness of red-black trees in maintaining balanced structures even under worst-case scenarios.

References

• Cormen, T. H., Leiserson, C. E., Rivest, R. L., & Stein, C. (2009). Introduction to Algorithms (3rd ed.). The MIT Press.
• Sedgewick, R., & Wayne, K. (2011). Algorithms (4th Edition). Addison-Wesley.