Binary Search for 3 in 2 4 8 9 1 3 6 by SakhinetiPraveena

Advanced/C++/BST.cpp

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+//SakhinetiPraveena
+//Given array 1 2 3 4 5, code for constructing a balanced binary search tree .
+#include<bits/stdc++.h>
+using namespace std;
+
+struct Node{
+    int data;
+    Node *left,*right;
+    Node(int val){
+        data=val;
+        left=NULL;
+        right=NULL;
+    }
+};
+// As the given array is sorted, to construct balanced BST the mid element is taken as root
+
+Node* ConstructBST(int arr[],int start, int end_){
+    if(start>end_) return NULL;
+    int mid=(start+end_)/2;
+    Node * root=new Node(arr[mid]);
+    root->left=ConstructBST(arr,start,mid-1);
+    root->right=ConstructBST(arr,mid+1,end_);
+    return root;
+}
+
+// function to print the inorder of BST
+void print_inorder(Node* root){
+    if(root==NULL) return;
+    if (root != NULL){
+        print_inorder(root->left);
+        cout << root->data <<" ";
+        print_inorder(root->right);
+    }
+}
+
+int main(){
+    int arr[5]={1,2,3,4,5};
+    Node* root=ConstructBST(arr,0,4);
+    cout<<"Inorder for the BST is :";
+    print_inorder(root);
+}

Intermediate/C++/BinarySearch.cpp

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+//SakhinetiPraveena
+//Binary Search for 3 in 2 4 8 9 1 3 6
+
+#include <bits/stdc++.h>
+using namespace std;
+
+void binarySearch(int arr[], int key,int n){
+    int s=0,e=n-1;                                    
+    while(s<=e){
+        int mid=(s+e)/2;
+        if(arr[mid]==key){
+            cout<<key<<" is found in the array.";
+            return;
+        }
+        else if(arr[mid]<key){
+            s=mid+1;
+        }
+        else e=mid-1;
+    }
+    cout<<key<<" is not found in the array.";
+}
+int main(){
+    int n=7;
+    int arr[7]={2, 4, 8, 9, 1, 3, 6};
+    sort(arr, arr + n);
+    binarySearch(arr,3,7);
+    return 0;
+}