Create a balanced BST using vector in C++ STL
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 3Input: 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:
- Firstly, we will sort the vector using the sort function.
- Now, get the Middle of the vector and make it root.
- Recursively do the same for the left half and the right half.
- Get the middle of the left half and make it the left child of the root created in step 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 Treeclass 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 vectorNode* 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 BSTvector<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 Codeint 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
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