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Thread: Test the equivalence of two binary trees
Started 1 month, 1 week ago by MindstorM
I want to make a java code which tests the equivalence of two binary trees . For trivial cases (tree size 0,1, ...) the code seems pretty obvious. However when the size is increasing, I can not determine an algorithm that returns me the correct results. Could you help me find this algo so that I can continue to code? Or give me links? On google I did not really find what I want by ...
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Forum: Software Development  Software Development - forum profile
Total authors: 2 authors
Total thread posts: 13 posts
Thread activity: no new posts during last week
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MindstorM replied 1 month, 1 week ago
In fact I have an expression such as E1 = -(A+B)*(C+D)/(C-D) E2 = (A*C + A*D + B*C + B*D) / (D – C) I have to build two trees with and create a method that determines if two trees are equivalent. For example A1.isEqual(Tree A2) should return true in this case.
Size: 369 bytes
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fellah replied 1 month, 1 week ago
When you say equity, you mean full equality, the 2 trees and structurally identical data? If so this should be a recursive algorithm in this style: Code: boolean estEgal(Node node1, Node node2) { if(node1 == null) { return (node2 == null); } if(node2 == null) { return false; } return (node1.value == node2.value) && isEqual(node1....
Size: 901 bytes
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fellah replied 1 month, 1 week ago
OK but it does not answer my questions: 1) How does your binary tree represents these formulas? What is their in the leaves? 2) What is the definition of "equivalent"?
Size: 266 bytes
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fellah replied 1 month, 1 week ago
So nodes contain operations and leaves contain constants. Already, it allows multiple representations of the same formula. In fact this representation form of binary tree is more suitable for RPN(Reverse Polish Notation). For me two binary trees are equivalent to two trees that contain exactly the same elements, but that does not respond to your statement (refer to the true representation ...
Size: 527 bytes
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MindstorM replied 1 month, 1 week ago
For E1 = -(A+B)*(C+D)/(C-D), for example, was this (sorry for the layout, this thing is capricious): Code: (*) / \ (-) (/) / / \ (+) (+) (+) / \ / \ / \ A B C D C D And just for the definition! Nothing in my course! Not to mention google
Size: 720 bytes
Customize:  Customize "<b>Reply 5</b>: Test the equivalence of two binary trees :: Software Development :: TechArena Community - Technology and Computer Hard..."

MindstorM replied 1 month, 1 week ago
In fact I have an expression such as E1 = -(A+B)*(C+D)/(C-D) E2 = (A*C + A*D + B*C + B*D) / (D ? C) I have to build two trees with and create a method that determines if two trees are equivalent. For example A1.isEqual(Tree A2) should return true in this case.
Size: 368 bytes
Customize:  Customize "<b>Reply 6</b>: Test the equivalence of two binary trees :: Software Development :: TechArena Community - Technology and Computer Hard..."

MindstorM replied 1 month, 1 week ago
I want to make a java code which tests the equivalence of two binary trees . For trivial cases (tree size 0,1, ...) the code seems pretty obvious. However when the size is increasing, I can not determine an algorithm that returns me the correct results. Could you help me find this algo so that I can continue to code? Or give me links? On google I did not really find what I want by ...
Size: 553 bytes
Customize:  Customize "<b>Reply 7</b>: Test the equivalence of two binary trees :: Software Development :: TechArena Community - Technology and Computer Hard..."

fellah replied 1 month, 1 week ago
When you say equity, you mean full equality, the 2 trees and structurally identical data? If so this should be a recursive algorithm in this style: Code: boolean estEgal(Node node1, Node node2) { if(node1 == null) { return (node2 == null); } if(node2 == null) { return false; } return (node1.value == node2.value) && isEqual(node1....
Size: 901 bytes
Customize:  Customize "<b>Reply 8</b>: Test the equivalence of two binary trees :: Software Development :: TechArena Community - Technology and Computer Hard..."

fellah replied 1 month, 1 week ago
OK but it does not answer my questions: 1) How does your binary tree represents these formulas? What is their in the leaves? 2) What is the definition of "equivalent"?
Size: 266 bytes
Customize:  Customize "<b>Reply 9</b>: Test the equivalence of two binary trees :: Software Development :: TechArena Community - Technology and Computer Hard..."

MindstorM replied 1 month, 1 week ago
For E1 = -(A+B)*(C+D)/(C-D), for example, was this (sorry for the layout, this thing is capricious): Code: (*) / \ (-) (/) / / \ (+) (+) (+) / \ / \ / \ A B C D C D And just for the definition! Nothing in my course! Not to mention google
Size: 720 bytes
Customize:  Customize "<b>Reply 10</b>: Test the equivalence of two binary trees :: Software Development :: TechArena Community - Technology and Computer Hard..."

 

Top contributing authors

Name
Posts
MindstorM
7
user's latest post:
Test the equivalence of two...
Published (2009-11-06 04:08:00)
I want to make a java code which tests the equivalence of two binary trees . For trivial cases (tree size 0,1, ...) the code seems pretty obvious. However when the size is increasing, I can not determine an algorithm that returns me the correct results. Could you help me find this algo so that I can continue to code? Or give me links? On google I did not really find what I want by searching this algo.
fellah
6
user's latest post:
Test the equivalence of two...
Published (2009-11-06 04:08:00)
When you say equity, you mean full equality, the 2 trees and structurally identical data? If so this should be a recursive algorithm in this style: Code: boolean estEgal(Node node1, Node node2) { if(node1 == null) { return (node2 == null); } if(node2 == null) { return false; } return (node1.value == node2.value) &amp;&amp; isEqual(node1.childLeft, node2.childLeft) &amp;&amp; isEqual(node1.childRight, node2.childRight); } And...

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