CSE331_HW4 - 1. a. 3 1 4 2 6 5 9 7 b. 4 1 6 2 5 7 9 2. a. C...

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1. a. 3 1 4 2 6 5 9 7 b. 4 1 6 2 5 9 7 2. a. C X G B Y P A
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b. This is the only way to make such a binary tree because the entire tree can only be drawn this way to represent both in-order and pre-order. For example for the right edge you can do it differently for just in-order but because it has to be the same as pre-order you have to put the node p on the right child of node g. Since in-order is starting from the left it ensures node P is last. c. The method I used was going node by node and fitting them in. I knew C would have to be the root node because it was the first in pre-order. Next I knew the farthest left node was going to have to be node B to satisfy in-order. I made X the parent node of B to also satisfy in-order and knew that my left most edge was complete because it satisfied both in and pre-order. Next I knew I had to place A and Y in the left side of the tree because C was my root node. The next part was easy because with pre-order there are more ways to write trees then in-order so I just started the next farthest left after X would be making A
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This note was uploaded on 03/31/2008 for the course CSE 331 taught by Professor M.mccullen during the Spring '08 term at Michigan State University.

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CSE331_HW4 - 1. a. 3 1 4 2 6 5 9 7 b. 4 1 6 2 5 7 9 2. a. C...

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