MHB Algorithm Type: Postorder Tree | Correct?

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The discussion centers around identifying the structure and computation of a postorder tree. The computation is suggested to be expressed as (2*3)+((4*2)-(1+5)), which is confirmed as correct. Participants clarify the distinction between a postorder tree and postorder traversal, with some confusion regarding the existence of left or right subtrees. The conversation also touches on the relationship between postorder trees and concepts of recursion and repetition. Overall, the focus remains on understanding the postorder tree structure and its computational implications.
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/ \

/ \

* -

/ \ / \

2 3 * +

/ \ / \

4 2 1 5


What type of algorithm is this I think the computation can be expressed as (2*3)+((4*2)-(1+5)). is this correct. I know its a postorder tree I just don't know if a left or right subtree exists Print root end. Does it anything to do with recursion or repetition
 
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           +

       /       \

      /         \

      *            -

     /  \       /     \

    2   3      *       +

             /  \     /  \

            4   2    1   5

barbara said:
What type of algorithm is this
This is not an algorithm.

barbara said:
I think the computation can be expressed as (2*3)+((4*2)-(1+5)).
Yes.

barbara said:
I know its a postorder tree
What is a postorder tree? I know what a postorder traversal is, but not sure about a postorder tree.

barbara said:
I just don't know if a left or right subtree exists Print root end.
I can't parse this sentence.

barbara said:
Does it anything to do with recursion or repetition
There is some connection.
 
The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...

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