Deriving time complexity of Binary tree and AVL tree, step
by step.
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Answer / tilak chandan
Lemma: A subtree rooted at node v has at least 2bh(v) − 1 internal nodes.
Proof of Lemma (by induction height):
Basis: h(v) = 0
If v has a height of zero then it must be null, therefore bh(v) = 0. So:
2bh(v) − 1 = 20 − 1 = 1 − 1 = 0
Inductive Step: v such that h(v) = k, has at least 2bh(v) − 1 internal nodes implies that v' such that h(v') = k+1 has at least 2bh(v') − 1 internal nodes.
Since v' has h(v') > 0 it is an internal node. As such it has two children each of which have a black-height of either bh(v') or bh(v')-1 (depending on whether the child is red or black, respectively). By the inductive hypothesis each child has at least 2bh(v') − 1 − 1 internal nodes, so v' has at least:
2bh(v') − 1 − 1 + 2bh(v') − 1 − 1 + 1 = 2bh(v') − 1
internal nodes.
Using this lemma we can now show that the height of the tree is logarithmic. Since at least half of the nodes on any path from the root to a leaf are black (property 4 of a red black tree), the black-height of the root is at least h(root)/2. By the lemma we get:
Therefore the height of the root is O(log(n)).
| Is This Answer Correct ? | 2 Yes | 1 No |
Answer / pudwallis gudzinsky
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| Is This Answer Correct ? | 5 Yes | 22 No |
Answer / haresh
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what the answer is.. But yah if there are girls in ur room i
will be happy striking with them lol
| Is This Answer Correct ? | 1 Yes | 19 No |
Answer / ssss ss
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strike with the girls available in the room.
| Is This Answer Correct ? | 6 Yes | 34 No |
Perform the functionality of 2-D array through 1-D array and in it the functions to be performed were: (1) Display the array in 2-D format (2) Display a particular element (3) Display a particular row (4) Display a particular column
Definition of priority queue was given. We have to implement the priority queue using array of pointers with the priorities given in the range 1..n. The array could be accessed using the variable top. The list corresponding to the array elements contains the items having the priority as the array index. Adding an item would require changing the value of top if it has higher priority than top. Extracting an item would require deleting the first element from the corresponding queue. The following class was given: class PriorityQueue { int *Data[100]; int top; public: void put(int item, int priority); // inserts the item with the given priority. int get(int priority); // extract the element with the given priority. int count(); // returns the total elements in the priority queue. int isEmpty(); // check whether the priority queue is empty or not. }; We had to implement all these class functions.
0 Answers Nagarro, Wollega University,
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0 Answers Jomo Kenyatta University,
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