Reference the following pseudocode for the next question: BSTadd(t, v) // from visualgo.net if insertion point is found create a new node if v < current node's value go left else go right 2.4. What would the tree look like after the following operations? Choose from the options below. BSTadd(t, 11) BSTadd(t, 9) BSTadd(t, 10) BSTremove(t, 8) 3 نیا 1 1 BSTremove(t, v) // from visualgo.net search for vif v is a leaf delete leafv else if v has 1 child bypass v else replace v with successor A 3 B 8 1 1

Database System Concepts
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Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
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# Binary Search Tree Operations

## Reference for Pseudocode

### Insertion
**BSTadd(t, v)**  
- From visualgo.net  
- If insertion point is found, create a new node.
- If `v <` current node's value, go left; otherwise, go right.

### Removal
**BSTremove(t, v)**  
- From visualgo.net  
- Search for `v`.
- If `v` is a leaf, delete the leaf.
- If `v` has one child, bypass `v`.
- Otherwise, replace `v` with its successor.

---

## Question

### 2.4. What would the tree look like after the following operations? Choose from the options below.

1. **BSTadd(t, 11)**
2. **BSTadd(t, 9)**
3. **BSTadd(t, 10)**
4. **BSTremove(t, 8)**

## Options

### Option A
- Root: `3`
- Left Child: `1`
- Right Subtree: 
  - Parent: `1`
  - Right Child: `9`
  - Left Child: `1`
  - Further right chain of `1`s leading to leaf `1`
  - End with leaf `9`

### Option B
- Root: `8`
- Left Child: `3`
- Further left chain: `1`
- Right Subtree: 
  - Parent: `1`
  - Repeating `1`s leading to leaf `1`
  - Left Child: `9`
  - Further down to leaf `1`
  
To answer, determine the correct structure based on the operations.
Transcribed Image Text:# Binary Search Tree Operations ## Reference for Pseudocode ### Insertion **BSTadd(t, v)** - From visualgo.net - If insertion point is found, create a new node. - If `v <` current node's value, go left; otherwise, go right. ### Removal **BSTremove(t, v)** - From visualgo.net - Search for `v`. - If `v` is a leaf, delete the leaf. - If `v` has one child, bypass `v`. - Otherwise, replace `v` with its successor. --- ## Question ### 2.4. What would the tree look like after the following operations? Choose from the options below. 1. **BSTadd(t, 11)** 2. **BSTadd(t, 9)** 3. **BSTadd(t, 10)** 4. **BSTremove(t, 8)** ## Options ### Option A - Root: `3` - Left Child: `1` - Right Subtree: - Parent: `1` - Right Child: `9` - Left Child: `1` - Further right chain of `1`s leading to leaf `1` - End with leaf `9` ### Option B - Root: `8` - Left Child: `3` - Further left chain: `1` - Right Subtree: - Parent: `1` - Repeating `1`s leading to leaf `1` - Left Child: `9` - Further down to leaf `1` To answer, determine the correct structure based on the operations.
The image contains two tree diagrams labeled "C" and "D." Each tree consists of nodes represented by circles containing numbers, connected by arrows indicating the direction of flow or relationship.

### Diagram C
- The root node has a value of 9.
- From the root, there are two primary branches:
  - The left branch leads to a node with the value 3, which further branches into a terminal node with the value 1.
  - The right branch from 9 leads to a node with the value 1. This node further splits into two branches:
    - One leading to another node with the value 1, which branches out into two terminal nodes both with the value 1.
    - The other directly leading to a terminal node with the value 1.

### Diagram D
- The root node has a value of 8.
- From the root, there are similarly two primary branches:
  - The left branch directly leads to a node with the value 3, which further branches into a terminal node with the value 1.
  - The right branch from 8 leads to a node with the value 1. This node branches into a node, again with the value 1, which itself branches into two terminal nodes each with the value 1.

Both diagrams showcase hierarchical tree structures with numeric nodes, illustrating potential paths or flows from root to terminal nodes.
Transcribed Image Text:The image contains two tree diagrams labeled "C" and "D." Each tree consists of nodes represented by circles containing numbers, connected by arrows indicating the direction of flow or relationship. ### Diagram C - The root node has a value of 9. - From the root, there are two primary branches: - The left branch leads to a node with the value 3, which further branches into a terminal node with the value 1. - The right branch from 9 leads to a node with the value 1. This node further splits into two branches: - One leading to another node with the value 1, which branches out into two terminal nodes both with the value 1. - The other directly leading to a terminal node with the value 1. ### Diagram D - The root node has a value of 8. - From the root, there are similarly two primary branches: - The left branch directly leads to a node with the value 3, which further branches into a terminal node with the value 1. - The right branch from 8 leads to a node with the value 1. This node branches into a node, again with the value 1, which itself branches into two terminal nodes each with the value 1. Both diagrams showcase hierarchical tree structures with numeric nodes, illustrating potential paths or flows from root to terminal nodes.
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