TADM2E 3.9 - Revision history

Matt: Undo revision 1012 by MattWars (talk)

2020-07-28T09:56:01Z

Undo revision 1012 by MattWars (talk)

MattWars at 07:16, 28 July 2020

2020-07-28T07:16:42Z

Matt: Undo revision 796 by EthanGamer (talk)

2020-07-23T13:30:47Z

Undo revision 796 by EthanGamer (talk)

EthanGamer: Blanked the page

2020-07-14T01:59:35Z

Blanked the page

Algowikiadmin: Recovering wiki

2014-09-11T18:13:03Z

Recovering wiki

New page

Since all the elements in S2 have keys larger than the keys of the elements in S1, those two trees can be subtrees of one tree whose root node will have a key larger than the largest key in S1 and smaller than the smallest key in S2. Then S2 will be the right subtree and S1 the left subtree. The easiest way to do this is to find the minimum element of S2 (or the maximum of S1) and put it as the root of the concatenated tree and than update its left and right pointer to point to the root of S1 and S2 respectively. The worst-case running time will be O(h), which is the time of retrieving the minimum (maximum) element of a tree.

[[Data-structures-TADM2E|Back to ''Data Structures'' Problems]]...

← Older revision		Revision as of 09:56, 28 July 2020
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−	Since all the elements in S2 have keys larger than the keys of the elements in S1, those two trees can be subtrees of one tree whose root node will have a key larger than the largest key in S1 and smaller than the smallest key in S2. Then S2 will be the right subtree and S6 the left subtree. The easiest way to do this is to find the minimum element of S2 (or the maximum of S1) and put it as the root of the concatenated tree and than update its left and right pointer to point to the root of S1 and S2 respectively. The worst-case running time will be O(h), which is the time of retrieving the minimum (maximum) element of a tree.	+	Since all the elements in S2 have keys larger than the keys of the elements in S1, those two trees can be subtrees of one tree whose root node will have a key larger than the largest key in S1 and smaller than the smallest key in S2. Then S2 will be the right subtree and S1 the left subtree. The easiest way to do this is to find the minimum element of S2 (or the maximum of S1) and put it as the root of the concatenated tree and than update its left and right pointer to point to the root of S1 and S2 respectively. The worst-case running time will be O(h), which is the time of retrieving the minimum (maximum) element of a tree.

	[[Data-structures-TADM2E\|Back to ''Data Structures'' Problems]]...		[[Data-structures-TADM2E\|Back to ''Data Structures'' Problems]]...

← Older revision		Revision as of 07:16, 28 July 2020
Line 1:		Line 1:
−	Since all the elements in S2 have keys larger than the keys of the elements in S1, those two trees can be subtrees of one tree whose root node will have a key larger than the largest key in S1 and smaller than the smallest key in S2. Then S2 will be the right subtree and S1 the left subtree. The easiest way to do this is to find the minimum element of S2 (or the maximum of S1) and put it as the root of the concatenated tree and than update its left and right pointer to point to the root of S1 and S2 respectively. The worst-case running time will be O(h), which is the time of retrieving the minimum (maximum) element of a tree.	+	Since all the elements in S2 have keys larger than the keys of the elements in S1, those two trees can be subtrees of one tree whose root node will have a key larger than the largest key in S1 and smaller than the smallest key in S2. Then S2 will be the right subtree and S6 the left subtree. The easiest way to do this is to find the minimum element of S2 (or the maximum of S1) and put it as the root of the concatenated tree and than update its left and right pointer to point to the root of S1 and S2 respectively. The worst-case running time will be O(h), which is the time of retrieving the minimum (maximum) element of a tree.

	[[Data-structures-TADM2E\|Back to ''Data Structures'' Problems]]...		[[Data-structures-TADM2E\|Back to ''Data Structures'' Problems]]...

← Older revision	Revision as of 13:30, 23 July 2020
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	+	Since all the elements in S2 have keys larger than the keys of the elements in S1, those two trees can be subtrees of one tree whose root node will have a key larger than the largest key in S1 and smaller than the smallest key in S2. Then S2 will be the right subtree and S1 the left subtree. The easiest way to do this is to find the minimum element of S2 (or the maximum of S1) and put it as the root of the concatenated tree and than update its left and right pointer to point to the root of S1 and S2 respectively. The worst-case running time will be O(h), which is the time of retrieving the minimum (maximum) element of a tree.

	+	[[Data-structures-TADM2E\|Back to ''Data Structures'' Problems]]...

← Older revision		Revision as of 01:59, 14 July 2020
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−	Since all the elements in S2 have keys larger than the keys of the elements in S1, those two trees can be subtrees of one tree whose root node will have a key larger than the largest key in S1 and smaller than the smallest key in S2. Then S2 will be the right subtree and S1 the left subtree. The easiest way to do this is to find the minimum element of S2 (or the maximum of S1) and put it as the root of the concatenated tree and than update its left and right pointer to point to the root of S1 and S2 respectively. The worst-case running time will be O(h), which is the time of retrieving the minimum (maximum) element of a tree.

−	~~[[Data-structures-TADM2E\|Back to ''Data Structures'' Problems]]...~~