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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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Answer:

hello some parts of your question is missing attached below is the missing part

answer :

A) Determine changes in the 50-mm dimensions

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Explanation:

<em>Given data</em> :

50-mm cube of graphite fiber reinforced polymer matrix

subjected to 125-KN force in direction 2,

direction 2 is perpendicular to fiber direction ( direction 1 ) and cube is constrained against expansion in direction 3

A) Determine changes in the 50-mm dimensions

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B) the stress required are evenly distributed

attached below is the detailed solution

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Read 2 more answers
A cooling system load is 96,000 BTUh sensible. How much chilled air is required to satisfy the load if the system is designed fo
Natalija [7]

Answer:

For 20^{\circ} - 5.556 lb/s

For 15^{\circ} - 7.4047 lb/s

Solution:

As per the question:

System Load = 96000 Btuh

Temperature, T = 20^{\circ}

Temperature rise, T' = 15^{\circ}

Now,

The system load is taken to be at constant pressure, then:

Specific heat of air, C_{p} = 0.24 btu/lb ^{\circ}F

Now, for a rise of 20^{\circ} in temeprature:

\dot{m}C_{p}\Delta T = 96000

\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 20} = 20000 lb/h = \frac{20000}{3600} = 5.556 lb/s

Now, for 15^{\circ}:

\dot{m}C_{p}\Delta T = 96000

\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 15} = 26666.667 lb/h = \frac{26666.667}{3600} = 7.4074 lb/s

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