The logic behind searching a node in a binary search tree is as follows:
Time complexity of searching a node in a binary search tree with N nodes : Worst-case time complexity is O(N), if the tree is linear; i.e to search in a binary search tree that is linear (unbalanced) all the nodes will have to be traversed and compared. Best-case time complexity is O(log N), if the tree is balanced, i.e for a balanced binary search tree the height would be log N and hence log N comparision before a node is found in the tree.
C++ : Searching a binary search tree. Implemented in C++11
#include<iostream>
#include<vector>
using namespace std;
class Node {
public:
int data;
Node * left;
Node * right;
Node(int x) : data(x), left(NULL), right(NULL)
{}
};
class Tree {
public:
vector<int> order;
/* The in-order traversal of a valid binary search tree is always in increasing order (sorted) */
void InOrder(Node * node) {
if (node != NULL) {
InOrder(node->left);
cout << node->data << " ";
order.push_back(node->data);
InOrder(node->right);
}
}
Node * SearchBST(Node * root, int data) {
if (root == NULL or root->data == data) {
return root;
} else if (root->data < data) {
return SearchBST(root->right, data);
} else if (root->data > data) {
return SearchBST(root->left, data);
}
return NULL;
}
};
int main() {
/* 6
/ \
4 9 */
Node * root1 = new Node(6);
root1->left = new Node(4);
root1->right = new Node(9);
Tree t;
int data = 9;
if (t.SearchBST(root1, data) != NULL) {
cout << "Node " << data << " found" << endl;
}
/* 5
/ \
1 7
\
8 */
Node * root2 = new Node(5);
root2->left = new Node(1);
root2->right = new Node(7);
root2->right->right = new Node(8);
data = 7;
if (t.SearchBST(root2, data) != NULL) {
cout << "Node " << data << " found" << endl;
}
data = 10;
if (t.SearchBST(root2, data) != NULL) {
cout << "Node " << data << " found" << endl;
}
return 0;
}
Output of validating a binary search tree implemented in C++11
Node 9 found.
Node 7 found.
Node 10 not found.