Binary search tree exact match - illustrated walkthrough

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A step-by-step illustration of Binary Search Tree (Exact Match) to help you walk through a series of operations. Illustration is accompanied by actual code with bold line indicating the current operation.

Technologie Business

Binary Search Tree
- Exact match
Illustrated walk through

Node structure
public class Node
{
public int Value;
public Node Left;
public Node Right;
}

Value

Left

Right

Find Exact Match
public Node FindExact(Node node, int target) {
if (node == null) {
return null;
}
if (node.Value == target) {
return node;
}
if (node.Value < target) {
return FindExact(node.Right, target);
}
else {
return FindExact(node.Left, target);
}
}

Base Case:
return when beyond a leaf
return a node found

Recursively search left or
right subtree

node

target

5
3

1

10

4

7

12

public Node FindExact( Node node, int target) {
if (node == null) {
return null;
}
if (node.Value == target) {
return node;
}
if (node.Value < target) {
return FindExact(node.Right, target);
}
else {
return FindExact(node.Left, target);
}
}

7

node

target

5
3

node == null
is false

1

10

4

7

12

public Node FindExact(Node node, int target) {
if (node == null) {
return null;
}
if (node.Value == target) {
return node;
}
if (node.Value < target) {
return FindExact(node.Right, target);
}
else {
return FindExact(node.Left, target);
}
}

7

node

target

5
3

5 == 7
is false

1

10

4

7

12

public Node FindExact(Node node, int target) {
if (node == null) {
return null;
}
if (node.Value == target) {
return node;
}
if (node.Value < target) {
return FindExact(node.Right, target);
}
else {
return FindExact(node.Left, target);
}
}

7

node

target

5
3

5<7
is true

1

10

4

7

12

public Node FindExact(Node node, int target) {
if (node == null) {
return null;
}
if (node.Value == target) {
return node;
}
if (node.Value < target) {
return FindExact(node.Right, target);
}
else {
return FindExact(node.Left, target);
}
}

7

node

target

5
3

1

7

10

4

7

12

Call Stack for 5
public Node FindExact(Node node, int target) {
if (node == null) {
return null;
}
if (node.Value == target) {
return node;
}
if (node.Value < target) {
return FindExact(node.Right, target);
}
else {
return FindExact(node.Left, target);
}
}

target

5
node

3

1

7

10

4

7

12
Call Stack for 10
Call Stack for 5

public Node FindExact( Node node, int target) {
if (node == null) {
return null;
}
if (node.Value == target) {
return node;
}
if (node.Value < target) {
return FindExact(node.Right, target);
}
else {
return FindExact(node.Left, target);
}
}

target

5
node

3

node == null
is false

1

7

10

4

7

12
Call Stack for 10
Call Stack for 5

public Node FindExact(Node node, int target) {
if (node == null) {
return null;
}
if (node.Value == target) {
return node;
}
if (node.Value < target) {
return FindExact(node.Right, target);
}
else {
return FindExact(node.Left, target);
}
}

target

5
node

3

1

10 == 7
is false

7

10

4

7

12
Call Stack for 10
Call Stack for 5

public Node FindExact(Node node, int target) {
if (node == null) {
return null;
}
if (node.Value == target) {
return node;
}
if (node.Value < target) {
return FindExact(node.Right, target);
}
else {
return FindExact(node.Left, target);
}
}

target

5
node

3

else (10 > 7)
is true

1

7

10

4

7

12
Call Stack for 10
Call Stack for 5

public Node FindExact(Node node, int target) {
if (node == null) {
return null;
}
if (node.Value == target) {
return node;
}
if (node.Value < target) {
return FindExact(node.Right, target);
}
else {
return FindExact(node.Left, target);
}
}

target

5
3

1

node

7

10

4

7

12
Call Stack for 7
Call Stack for 10
Call Stack for 5

public Node FindExact( Node node, int target) {
if (node == null) {
return null;
}
if (node.Value == target) {
return node;
}
if (node.Value < target) {
return FindExact(node.Right, target);
}
else {
return FindExact(node.Left, target);
}
}

node == null
is false

target

5
3

1

node

7

10

4

7

12
Call Stack for 7
Call Stack for 10
Call Stack for 5

public Node FindExact(Node node, int target) {
if (node == null) {
return null;
}
if (node.Value == target) {
return node;
}
if (node.Value < target) {
return FindExact(node.Right, target);
}
else {
return FindExact(node.Left, target);
}
}

7 == 7
is true
3

1

node

7

target

5
10

4

7

12
Call Stack for 7
Call Stack for 10
Call Stack for 5

public Node FindExact(Node node, int target) {
if (node == null) {
return null;
}
if (node.Value == target) {
return node;
}

We found an
exact match!

if (node.Value < target) {
return FindExact(node.Right, target);
}
else {
return FindExact(node.Left, target);
}
}

target

5
3

7

10
Call Stack for 7

1

node

4

7

12
Call Stack for 10
Call Stack for 5

public Node FindExact(Node node, int target) {
if (node == null) {
return null;
}
if (node.Value == target) {
return node;
}
if (node.Value < target) {
return FindExact(node.Right, target);
}
else {
return FindExact(node.Left, target);
}
}

7

target

5
node

3

1

7

10

4

7

12
Call Stack for 10

7
Call Stack for 5
public Node FindExact(Node node, int target) {
if (node == null) {
return null;
}
if (node.Value == target) {
return node;
}
if (node.Value < target) {
return FindExact(node.Right, target);
}
else {
return FindExact(node.Left, target);
}
}

target

5

node
3

1

7

10

4

7

12

Call Stack for 5
public Node FindExact(Node node, int target) {
if (node == null) {
return null;
}
if (node.Value == target) {
return node;
}
if (node.Value < target) {
return FindExact(node.Right, target);
}
else {
return FindExact(node.Left, target);
}
}

7

target

5
3

1

7

10

4

7

12

public Node FindExact(Node node, int target) {
if (node == null) {
return null;
}

The caller of the
if (node.Value ==
FindExactreturn node;
receives
node 7}

target) {

if (node.Value < target) {
return FindExact(node.Right, target);
}
else {
return FindExact(node.Left, target);
}
}

7

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Given a newly created Binary Search Tree with the following numerical key input sequence (from left to right): 9, 4, 12, 7, 5, 2, 20, 14, 11, 13, 19, 16, 17, 42, 24 a.) We are asked to add a function Position lowestKey(const Position& v) const to the class SearchTree. Function prototype: Position lowestKey(const Position& v); input argument: position of the starting node in the binary search tree to calculate from. For the whole tree this would be the position for the root of the tree. return value: position of the node having the lowest key value. i. Describe strategy with reasonings and examples to convince that it will work. ii. Implement the function using C++ and show the source code listing. b) We are also asked to add a function Position highestKey(const Position& v) const to the class SearchTree. Function prototype: Position highestKey(const Position& v) const; input argument: position of the node in the binary search tree to calculate from. For the whole tree this would be the position for the root of the tree. return value: position of the node having the highest key value. i. Describe strategy with reasonings and examples to convince that it will work. ii. Implement the function using C++ and show the source code listing. Solution #include #include /* A binary tree node has data, pointer to left child and a pointer to right child */ struct node { int data; struct node* left; struct node* right; }; /* Helper function that allocates a new node with the given data and NULL left and right pointers. */ struct node* newNode(int data) { struct node* node = (struct node*) malloc(sizeof(struct node)); node->data = data; node->left = NULL; node->right = NULL; return(node); } /* Give a binary search tree and a number, inserts a new node with the given number in the correct place in the tree. Returns the new root pointer which the caller should then use (the standard trick to avoid using reference parameters). */ struct node* insert(struct node* node, int data) { /* 1. If the tree is empty, return a new, single node */ if (node == NULL) return(newNode(data)); else { /* 2. Otherwise, recur down the tree */ if (data <= node->data) node->left = insert(node->left, data); else node->right = insert(node->right, data); /* return the (unchanged) node pointer */ return node; } } /* Given a non-empty binary search tree, return the minimum data value found in that tree. Note that the entire tree does not need to be searched. */ int minValue(struct node* node) { struct node* current = node; /* loop down to find the leftmost leaf */ while (current->left != NULL) { current = current->left; } return(current->data); } /* Driver program to test sameTree function*/ int main() { struct node* root = NULL; root = insert(root, 4); insert(root, 2); insert(root, 1); insert(root, 3); insert(root, 6); insert(root, 5); printf(\"\ Minimum value in BST is %d\", minValue(root)); getchar(); return 0; } #include #include /* A binary tree node has data, pointer to left child .

Given a newly created Binary Search Tree with the following numerica.pdf

hadpadrrajeshh

package edu.ser222.m03_02; /** * A binary search tree based implementation of a symbol table. * * Completion time: (your completion time) * * @author (your name), Sedgewick, Acuna * @version (version) */ import java.util.Collections; import java.util.LinkedList; import java.util.NoSuchElementException; import java.util.Queue; public class CompletedBST<Key extends Comparable<Key>, Value> implements BST<Key, Value> { private Node<Key, Value> root; @Override public int size() { return size(root); } private int size(Node x) { if (x == null) return 0; else return x.N; } @Override public Value get(Key key) { Node<Key, Value> iter = root; while(iter != null) { int cmp = key.compareTo(iter.key); if (cmp < 0) iter = iter.left; else if (cmp > 0) iter = iter.right; else return iter.val; } return null; } private Value get(Node<Key, Value> x, Key key) { // Return value associated with key in the subtree rooted at x; // return null if key not present in subtree rooted at x. if (x == null) return null; int cmp = key.compareTo(x.key); if (cmp < 0) return get(x.left, key); else if (cmp > 0) return get(x.right, key); else return x.val; } @Override public void put(Key key, Value val) { root = put(root, key, val); } private Node put(Node<Key, Value> x, Key key, Value val) { if (x == null) return new Node(key, val, 1); int cmp = key.compareTo(x.key); if (cmp < 0) x.left = put(x.left, key, val); else if (cmp > 0) x.right = put(x.right, key, val); else x.val = val; x.N = size(x.left) + size(x.right) + 1; return x; } @Override public Key min() { if(root == null) throw new NoSuchElementException(); return min(root).key; } private Node<Key, Value> min(Node x) { if (x.left == null) return x; return min(x.left); } @Override public Key max() { if(root == null) throw new NoSuchElementException(); return max(root).key; } private Node<Key, Value> max(Node x) { if (x.right == null) return x; return max(x.right); } @Override public Key floor(Key key) { if(root == null) throw new NoSuchElementException(); Node<Key, Value> x = floor(root, key); if (x == null) return null; return x.key; } private Node<Key, Value> floor(Node<Key, Value> x, Key key) { if (x == null) return null; int cmp = key.compareTo(x.key); if (cmp == 0) return x; if (cmp < 0) return floor(x.left, key); Node<Key, Value> t = floor(x.right, key); if (t != null) return t; else return x; } @Override public Key select(int k) { return select(root, k).key; } private Node<Key, Value> select(Node x, int k) { if (x == null) return null; int t = size(x.left); if (t > k) return select(x.left, k); else if (t < k) return select(x.right, k-t-1); else return x; } @Override public int rank(Key key) { return rank(key, root); } private int rank(Key key, Node<Key, Value> x) { // Return number of keys less than x.key in the subtree rooted at x. if (x == null) return 0; int cmp = key.compareTo(x.key); if (cmp < 0) return rank(key, x.left); else if (cmp > 0) return 1 + size(x.left) + rank(key, x.right); else return size(x.left); } @O.

package edu-ser222-m03_02- --- - A binary search tree based impleme.docx

farrahkur54

Help I keep getting the same error when running a code. Below is the code and the instructions about the code. import java.util.Arrays; public class Album implements Comparable { private final int id; private final String[] artists; private final int numSongs; public Album(int id, String title, String[] artists, int numSongs) { this.id = id; this.artists = new String[]{Arrays.toString(artists)}; this.numSongs = numSongs; } public int getId() { return id; } @Override public int compareTo(Album other) { return Integer.compare(this.numSongs, other.numSongs); } @Override public String toString() { StringBuilder artistNames = new StringBuilder(); for (String artist : artists) { artistNames.append(artist).append(", "); } artistNames.delete(artistNames.length() - 2, artistNames.length()); return "ID: " + id + " -- " + numSongs + " songs -- [" + artistNames + "]"; } } import java.util.Random; public class DoublyLinkedList { private static class Node { T data; Node prev; Node next; Node(T data) { this.data = data; } } private Node head; private Node tail; private int size; public DoublyLinkedList() { head = null; tail = null; size = 0; } public Node append(T data) { Node newNode = new Node<>(data); if (head == null) { head = newNode; } else { tail.next = newNode; newNode.prev = tail; } tail = newNode; size++; return newNode; } public Node insert(int location, T data) { if (location < 0 || location > size) { throw new IllegalArgumentException("Invalid location"); } Node newNode = new Node<>(data); if (location == 0) { newNode.next = head; if (head != null) { head.prev = newNode; } head = newNode; if (tail == null) { tail = newNode; } } else if (location == size) { tail.next = newNode; newNode.prev = tail; tail = newNode; } else { Node current = head; for (int i = 0; i < location; i++) { current = current.next; } newNode.prev = current.prev; newNode.next = current; current.prev.next = newNode; current.prev = newNode; } size++; return newNode; } public Node delete(int location) { if (location < 0 || location >= size) { throw new IllegalArgumentException("Invalid location"); } Node deletedNode; if (location == 0) { deletedNode = head; head = head.next; if (head != null) { head.prev = null; } else { tail = null; } } else if (location == size - 1) { deletedNode = tail; tail = tail.prev; tail.next = null; } else { Node current = head; for (int i = 0; i < location; i++) { current = current.next; } deletedNode = current; current.prev.next = current.next; current.next.prev = current.prev; } size--; return deletedNode; } public int getIndex(T data) { Node current = head; for (int i = 0; i < size; i++) { if (current.data.equals(data)) { return i; } current = current.next; } return -1; } @Override public String toString() { StringBuilder result = new StringBuilder(); Node current = head; while (current != null) { result.append(current.data).append(" -> "); current = current.next; } result.append("NULL"); return result.toString(); } public Node shuffle() { Random r.

Help I keep getting the same error when running a code. Below is the.pdf

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Please change this method to recursive method. public String postfix() { final ExpressionNodes nodes = new ExpressionNodes(); for (String element : expression.split(\" \")) { if (isOperator(element)) { String rightNode = nodes.pop(); String leftNode = nodes.pop(); nodes.push((leftNode + \" \" + rightNode + \" \" + element)); } else { nodes.push((element)); } } return nodes.pop(); } Solution public class Main{ public String postfix(int count) final ExpressiononNodes nodes = new ExpressiononNodes (); for( String element : expression split(\" \")) { if (Is Operator(int element)) { String rightNode = nodes.pop(); String leftNode = nodes.pop(); nodes.push((leftNode + \" \" + rightNode + \" \" +element)); } else { nodes.push ((element)); }} if (count <=0) { return nodes.pop(); } else { postfix(count - 1); }} Public static void Main(String[] args) { Main n = new Main(); n.postfix(element); }.

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ajoy21

Java code to find the closest pair using divide and conquer algorithm. Solution The Brute force solution is O(n^2), compute the distance between each pair and return the smallest. But Devide and conquer is better and completes in O(nLogn) time. Please find below the code: import java.util.*; // compile: javac ClosestPair.java // run: Java ClosestPair 500 // Here 500 is some randomly generated set of input points. /* The algorithm is : Input: An array of n points P[] Output: The smallest distance between two points in the given array. As a pre-processing step, input array is sorted according to x coordinates. 1) Find the middle point in the sorted array, we can take P[n/2] as middle point. 2) Divide the given array in two halves. The first subarray contains points from P[0] to P[n/2]. The second subarray contains points from P[n/2+1] to P[n-1]. 3) Recursively find the smallest distances in both subarrays. Let the distances be dl and dr. Find the minimum of dl and dr. Let the minimum be d. */ public class ClosestPair { public static class Point { public final double x; public final double y; public Point(double x, double y) { this.x = x; this.y = y; } public String toString() { return \"(\" + x + \", \" + y + \")\"; } } public static class Pair { public Point point1 = null; public Point point2 = null; public double distance = 0.0; public Pair() { } public Pair(Point point1, Point point2) { this.point1 = point1; this.point2 = point2; calcDistance(); } public void update(Point point1, Point point2, double distance) { this.point1 = point1; this.point2 = point2; this.distance = distance; } public void calcDistance() { this.distance = distance(point1, point2); } public String toString() { return point1 + \"-\" + point2 + \" : \" + distance; } } public static double distance(Point p1, Point p2) { double xdist = p2.x - p1.x; double ydist = p2.y - p1.y; return Math.hypot(xdist, ydist); } public static Pair bruteForce(List points) { int numPoints = points.size(); if (numPoints < 2) return null; Pair pair = new Pair(points.get(0), points.get(1)); if (numPoints > 2) { for (int i = 0; i < numPoints - 1; i++) { Point point1 = points.get(i); for (int j = i + 1; j < numPoints; j++) { Point point2 = points.get(j); double distance = distance(point1, point2); if (distance < pair.distance) pair.update(point1, point2, distance); } } } return pair; } public static void sortByX(List points) { Collections.sort(points, new Comparator() { public int compare(Point point1, Point point2) { if (point1.x < point2.x) return -1; if (point1.x > point2.x) return 1; return 0; } } ); } public static void sortByY(List points) { Collections.sort(points, new Comparator() { public int compare(Point point1, Point point2) { if (point1.y < point2.y) return -1; if (point1.y > point2.y) return 1; return 0; } } ); } public static Pair divideAndConquer(List points) { List pointsSortedByX = new ArrayList(points); sortByX(pointsSortedByX); List pointsSortedByY = new ArrayList(points); sortByY.

Java code to find the closest pair using divide and conquer algorith.pdf

fashioncollection2

Given interval is `[0,2pi]` . And given identity is `secx+tanx=1` which we can write as `1/(cosx)+(sinx)/(cosx)=1` òr` , `(1+sinx)/(cosx)=1` or, `(1+sinx)=cosx` or, `1+sinx-cosx=0` or, `1+2sin(x/2)cos(x/2)-(1-2sin^2(x/2))=0` or, `1+2sin(x/2)cos(x/2)-1+2sin^2(x/2)=0` or, `2sin(x/2)cos(x/2)+2sin^2(x/2)=0` or, `2sin(x/2)[cos(x/2)+sin(x/2)]=0` òr,` `2sin(x/2)=0` and `cos(x/2)+sin(x/2)=0` . Now, `2sin(x/2)=0` `rArrsin(x/2)=0` `rArrsin(x/2)=sin(npi)` , n=0,1,2. As per the need of our problem. `rArr x/2=npi` or, `x=2npi` for n=0,1 as our interval is `[0,2pi]` . So, `x=0` or, `x=2pi` . Now, if `cos(x/2)+sin(x/2)=0` `rArr1+(sin(x/2))/cos(x/2)=0` `rArrtan(x/2)=-1` or, `tan(x/2)=tan(pi/2+pi/4)` or, `x/2=(pi/2+pi/4)` or, `x=pi+pi/2` or, `x=3pi/2` . Clearly, `x=0` and `x=2pi` are the solutions satisfying our given identity over the given interval. Solution Given interval is `[0,2pi]` . And given identity is `secx+tanx=1` which we can write as `1/(cosx)+(sinx)/(cosx)=1` òr` , `(1+sinx)/(cosx)=1` or, `(1+sinx)=cosx` or, `1+sinx-cosx=0` or, `1+2sin(x/2)cos(x/2)-(1-2sin^2(x/2))=0` or, `1+2sin(x/2)cos(x/2)-1+2sin^2(x/2)=0` or, `2sin(x/2)cos(x/2)+2sin^2(x/2)=0` or, `2sin(x/2)[cos(x/2)+sin(x/2)]=0` òr,` `2sin(x/2)=0` and `cos(x/2)+sin(x/2)=0` . Now, `2sin(x/2)=0` `rArrsin(x/2)=0` `rArrsin(x/2)=sin(npi)` , n=0,1,2. As per the need of our problem. `rArr x/2=npi` or, `x=2npi` for n=0,1 as our interval is `[0,2pi]` . So, `x=0` or, `x=2pi` . Now, if `cos(x/2)+sin(x/2)=0` `rArr1+(sin(x/2))/cos(x/2)=0` `rArrtan(x/2)=-1` or, `tan(x/2)=tan(pi/2+pi/4)` or, `x/2=(pi/2+pi/4)` or, `x=pi+pi/2` or, `x=3pi/2` . Clearly, `x=0` and `x=2pi` are the solutions satisfying our given identity over the given interval..

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