The binary search tree is a binary tree where each node has at most two children, and the values of the left child are smaller than that of the parent node, while the values of the right child are greater than that of the parent node.
Given the following are the formation of a binary search tree for the words: {mathematics, physics, geography, zoology, meteorology, geology, psychology, and chemistry}.
Using alphabetical order, we will show the insertion of each word into the tree.
Then Insertion of Words into Binary Search Tree:
First Inserting the root nodeThe first word in the list is mathematics. Thus, it becomes the root of the binary search tree.
Then Inserting the second nodeNext, we insert physics. Physics is less than mathematics, so it is inserted as the left child of mathematics.
Similarly, Inserting the third next word in the list is geography.
As geography is greater than mathematics, it is inserted as the right child of mathematics.
And Inserting the fourth nodeThe next word in the list is zoology. Since zoology is less than physics, it becomes the left child of physics.
Then the fifth nodeThe next word in the list is meteorology. As meteorology is greater than mathematics but less than geography, it becomes the right child of mathematics.
Then the sixth nodeThe next word in the list is geology. As geology is less than geography, it becomes the left child of geography. (See diagram below)
As we can see that chemistry is less than mathematics but greater than physics, it becomes the left child of mathematics.
Hence, we get the following Binary Search Tree: detailed explanation is given above.
To know more about binary search visit:
brainly.com/question/33329263
#SPJ4
Binary search tree is a data structure used for storing data in an ordered manner. It allows fast search and insertion of new elements in a way that is similar to searching through a sorted list.
A Binary search tree (BST) is a binary tree where each node has values that are greater or equal to its left child and less or equal to its right child. It is used for searching and sorting operations. It provides fast search operations due to the fact that each comparison eliminates about half of the remaining nodes in the search space. The given words {chemistry, geography, geology, mathematics, meteorology, physics, psychology, zoology} in alphabetical order form the following binary search tree: mathematics physics zoology geography meteorology psychology chemistry geology. The first word, mathematics, is inserted at the root of the tree. Next, physics is inserted to the right of mathematics. Zoology is then inserted to the left of physics. Geography is inserted to the left of mathematics. Meteorology is inserted to the right of physics. Psychology is inserted to the left of chemistry. Finally, chemistry is inserted to the right of psychology.
Know more about Binary search, here:
https://brainly.com/question/13143459
#SPJ11
a groundwater contains 1.2 mg/l of iron as fe3 . what ph is required to precipitate all but 0.2 mg/l of the iron at 25 c.
The pH required to precipitate all but 0.2 mg/l of the iron in the groundwater at 25°C is approximately 9.0.
To understand why a pH of approximately 9.0 is required to precipitate most of the iron, we need to consider the solubility of iron at different pH levels. Iron in groundwater is usually present as Fe3+ ions. At a pH below 9.0, Fe3+ ions tend to remain soluble in water. However, when the pH rises above 9.0, hydroxide ions (OH-) are present in sufficient concentrations to react with the Fe3+ ions.
This reaction forms insoluble iron hydroxide (Fe(OH)3), which precipitates out of the solution. The solubility of Fe(OH)3 is very low at a pH above 9.0, resulting in the removal of most of the iron from the water. Therefore, adjusting the pH to around 9.0 enables the precipitation of iron, reducing its concentration in the groundwater.
Learn more about pH:
brainly.com/question/2288405
#SPJ11
1. Convert C to assembly. Assume all variables are integers, and initialized before these code blocks. Assume that there is code before and after these blocks, and that each part is independent from each other. a. Assume a is stored in RO, b is stored in R1, and is stored in R2. a = 1; b = 6; c = a + b; a b = a - c; c++; a <<= 3; b. Assume a is stored in RO, b is stored in R1, and c is stored in R2. if (a < b) a + C; } else! a - C; C. } Assume a is stored in RO, b is stored in R1, and c is stored in R2, x is stored in R3. for (x 10; x > 0; --) { C = C + x; a = a + c; b = a + x; } d. Assume a is stored in RO, b is stored in R1, and c is stored in R2. if ((a < 2 && a >= b) 11 (b != c)) { a += 2; } else! а = b;
Explanation:
Here's the conversion of the given C code blocks into assembly code, assuming the variables are stored in the specified registers:
a.
```assembly
li $r0, 1 # a = 1
li $r1, 6 # b = 6
add $r2, $r0, $r1 # c = a + b
sub $r0, $r0, $r2 # a = a - c
addi $r2, $r2, 1 # c++
sll $r0, $r0, 3 # a <<= 3
```
b.
```assembly
slt $r3, $r0, $r1 # if (a < b)
beqz $r3, else_label # jump to else_label if a >= b
add $r0, $r0, $r2 # a + c
j end_label # jump to end_label
else_label:
sub $r0, $r0, $r2 # a - c
end_label:
```
c.
```assembly
li $r3, 10 # x = 10
loop:
bgtz $r3, loop_end # exit loop if x <= 0
add $r2, $r2, $r3 # C = C + x
add $r0, $r0, $r2 # a = a + C
add $r1, $r0, $r3 # b = a + x
subi $r3, $r3, 1 # x--
j loop # jump to loop
loop_end:
```
d.
```assembly
slt $r3, $r0, 2 # if (a < 2)
beqz $r3, else_label # jump to else_label if a >= 2
slt $r3, $r0, $r1 # if (a >= b)
bnez $r3, else_label # jump to else_label if a < b
sub $r0, $r0, $r2 # a = b
j end_label # jump to end_label
else_label:
addi $r0, $r0, 2 # a += 2
end_label:
```
In the above assembly code, the instructions assume that the register `$r0` corresponds to variable `a`, `$r1` corresponds to variable `b`, and `$r2` corresponds to variable `c`. The register `$r3` is used for temporary operations or loop variables.
