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Nana76 [90]
3 years ago
10

Draw the sequence of BSTs that results when you insert the keys E, A, S, Y, Q, U, E, S, T, I, O, N, in that order into an initia

lly empty tree. And draw the sequence of BSTs that results when you delete the keys from the tree one by one in the order they were inserted. (Note: here your BST should allow duplicate keys).

Engineering
1 answer:
jek_recluse [69]3 years ago
8 0

Answer:

answer is attached

Explanation:

An important special kind of binary tree is the binary search tree (BST). In a BST, each node stores some information including a unique key value, and perhaps some associated data. A binary tree is a BST iff, for every node n in the tree:

All keys in n's left subtree are less than the key in n, and

all keys in n's right subtree are greater than the key in n.

Note: if duplicate keys are allowed, then nodes with values that are equal to the key in node n can be either in n's left subtree or in its right subtree (but not both). In these notes, we will assume that duplicates are not allowed.

Here are some BSTs in which each node just stores an integer key:

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ear shaft.3. Chapter 12 –Loading on Spur Gears: A 26-tooth pinion rotating at a uniform 1800 rpm meshes with a 55-tooth gear in
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Answer:

The bending stress is 502.22 MPa

Explanation:

The diameter of the pinion is equal to:

d_{p} =mN_{p}

Where

m = module = 5

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d_{p} =5*26=130mm = 0.13 m

The pitch line velocity is equal to:

V_{t} =\frac{d_{p}*2*\pi  *w_{p} }{120}

Where

wp = speed of the pinion = 1800 rpm

V_{t} =\frac{0.13*2*\pi *1800}{120} =12.25m/s

The factor B is equal to:

B=\frac{(12-Q_{v})^{2/3}  }{4} , if Q_{v} =10\\B=\frac{(12-10)^{2/3} }{4} =0.396

The factor A is equal to:

A = 50 + 56*(1 - B) = 50 + 56*(1-0.396) = 83.82

The dynamic factor is:

K_{v} =(\frac{A}{A+\sqrt{200V_{t} } } )^{B} \\K_{v}=(\frac{83.82}{83.82+\sqrt{200*12.25} } )^{0.396} =0.832

The geometry bending factor at 20°, the application factor Ka, load distribution factor Km, the size factor Ks, the rim thickness factor Kb and Ki the idler factor can be obtained from tables

JR = 0.41

Ka = 1

Kb = 1

Ks = 1

Ki = 1.42

Km = 1.7

The diametrical pitch is equal to:

P_{d} =\frac{1}{m} =\frac{1}{5} =0.2mm^{-1}

The bending stress is equal to:

\sigma =\frac{W_{t}P_{d}K_{a}K_{m}K_{s}K_{b}K_{i} }{FJ_{g}K_{v}}  \\\sigma =\frac{22000*0.2*1*1.7*1*1*1.42}{62*0.41*0.832} =502.22MPa

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Answer:

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