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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| Is This Answer Correct ? | 1 Yes | 19 No |
Answer / ssss ss
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| Is This Answer Correct ? | 6 Yes | 34 No |
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readers and writers problem
C Code 
