Skip to content
Related Articles

Related Articles

Create a balanced BST using vector in C++ STL

View Discussion
Improve Article
Save Article
  • Last Updated : 29 Dec, 2021

Given an unsorted vector arr, the task is to create a balanced binary search tree using the elements of the array.

Note: There can be more than one balanced BST. Forming any is acceptable

Examples:  

Input: arr[] = { 2, 1, 3}
Output: 2 1 3
Explanation: The tree formed is show below. The preorder traversal is 2 1 3
     2
   /  \
 1    3  

Input: arr[] = {4, 3, 1, 2}
Output: 2 1 3 4
Explanation: The tree formed is
        2
      /  \
    1    3
           \
            4
Another possible option can provide preorder traversal is 3 2 1 4

 

Approach: To solve this problem, follow the below steps:

  1. Firstly, we will sort the vector using the sort function.
  2. Now, get the Middle of the vector and make it root.
  3. Recursively do the same for the left half and the right half.
    1. Get the middle of the left half and make it the left child of the root created in step 2.
    2. Get the middle of the right half and make it the right child of the root created in step 2.

Below is the implementation of the above approach:

C++




// C++ program to print BST in given range
#include <bits/stdc++.h>
using namespace std;
  
// Node of Binary Search Tree
class Node {
public:
    Node* left;
    int data;
    Node* right;
  
    Node(int d)
    {
        data = d;
        left = right = NULL;
    }
};
  
// A function that constructs Balanced
// Binary Search Tree from a vector
Node* createBST(vector<int> v, int start, 
                int end)
{
    sort(v.begin(), v.end());
  
    // Base Case
    if (start > end)
        return NULL;
  
    // Get the middle element and make it root
    int mid = (start + end) / 2;
    Node* root = new Node(v[mid]);
  
    // Recursively construct the left subtree
    // and make it left child of root
    root->left = createBST(v, start, mid - 1);
  
    // Recursively construct the right subtree
    // and make it right child of root
    root->right = createBST(v, mid + 1, end);
  
    return root;
}
  
vector<int> preNode, vec;
  
// A utility function to print
// preorder traversal of BST
vector<int> preOrder(Node* node)
{
    // Root Left Right
    if (node == NULL) {
        return vec;
    }
    preNode.push_back(node->data);
    preOrder(node->left);
    preOrder(node->right);
    return preNode;
}
  
// Driver Code
int main()
{
    vector<int> v = { 4, 3, 1, 2 };
    Node* root = createBST(v, 0, v.size() - 1);
  
    vector<int> ans = preOrder(root);
    for (auto i : ans) {
        cout << i << " ";
    }
    return 0;
}


Output

PreOrder Traversal of constructed BST 
4 2 1 3 6 5 7 

Time Complexity: O(N * logN)
Auxiliary Space: O(N) to create the tree


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!