C++ Program to Search for an Element in a Binary Search Tree

In this program we need to. Implement binary search to find the existence of a search sequence in a binary search tree. The worst case time complexity of Binary search is O(n) but for the average case O(log(n)).

Algorithm

Begin
Construct binary search tree for the given unsorted data array by inserting data into tree one by one.
Take the input of data to be searched in the BST.
Now starting from the root node, compare the data with data part of the node.
if data < temp->d, move the temp pointer to the left child.
if data > temp->d move the temp pointer to the right child.
if data = temp->d print the tree depth where it is found and return to main.
End

Example Code

#include<iostream>
using namespace std;
struct node {
int d;
node *left;
node *right;
};
node* CreateNode(int d) {
node *newnode = new node;
newnode->d = d;
newnode->left = NULL;
newnode->right = NULL;
return newnode;
}
node* InsertIntoTree(node* root, int d) {
node *temp = CreateNode(d);
node *t = new node;
t = root;
if(root == NULL)
root = temp;
else {
while(t != NULL) {
if(t->d < d) {
if(t->right == NULL) {
t->right = temp;
break;
}
t = t->right;
} else if(t->d > d) {
if(t->left == NULL) {
t->left = temp;
break;
}
t = t->left;
}
}
}
return root;
}
void Search(node *root, int d) {
int depth = 0;
node *temp = new node;
temp = root;
while(temp != NULL) {
depth++;
if(temp->d == d) {
cout<<"\nitem found at depth: "<<depth;
return;
} else if(temp->d > d)
temp = temp->left;
else
temp = temp->right;
}
return;
}
int main() {
char ch;
int n, i, a[10] = {93, 53, 45, 2, 7, 67, 32, 26, 71, 76};
node *root = new node;
root = NULL;
for (i = 0; i < 10; i++)
root = InsertIntoTree(root, a[i]);
up:
cout<<"\nEnter the Element to be searched: ";
cin>>n;
Search(root, n);
cout<<"\n\n\tDo you want to search more...enter choice(y/n)?";
cin>>ch;
if(c == 'y' || c == 'Y')
goto up;
return 0;
}

Output

Enter the Element to be searched: 26
item found at depth: 7
Do you want to search more...enter choice(y/n)?
Enter the Element to be searched: 1
Do you want to search more...enter choice(y/n)?