Balance a Binary Search Tree Last Updated : 23 Jul, 2025 Comments Improve Suggest changes Like Article Like Report Try it on GfG Practice Given a BST (Binary Search Tree) that may be unbalanced, the task is to convert it into a balanced BST that has the minimum possible height.Examples: Input: Output: Explanation: The above unbalanced BST is converted to balanced with the minimum possible height.Input: Output: Explanation: The above unbalanced BST is converted to balanced with the minimum possible height.Approach:The idea is to store the elements of the tree in an array using inorder traversal. Inorder traversal of a BST produces a sorted array. Once we have a sorted array, recursively construct a balanced BST by picking the middle element of the array as the root for each subtree.Follow the steps below to solve the problem:Traverse given BST in inorder and store result in an array. Note that this array would be sorted as inorder traversal of BST always produces sorted sequence.Build a balanced BST from the above created sorted array using the recursive approach discussed in Sorted Array to Balanced BST. C++ //Driver Code Starts // C++ program to convert a left unbalanced BST to // a balanced BST #include <iostream> #include <vector> #include <queue> using namespace std; class Node { public: int data; Node* left; Node* right; Node(int value) { data = value; left = nullptr; right = nullptr; } }; //Driver Code Ends // Inorder traversal to store elements of the // tree in sorted order void storeInorder(Node* root, vector<int>& nodes) { if (root == nullptr) return; // Traverse the left subtree storeInorder(root->left, nodes); // Store the node data nodes.push_back(root->data); // Traverse the right subtree storeInorder(root->right, nodes); } // Function to build a balanced BST from a sorted array Node* buildBalancedTree(vector<int>& nodes, int start, int end) { // Base case if (start > end) return nullptr; // Get the middle element and make it the root int mid = (start + end) / 2; Node* root = new Node(nodes[mid]); // Recursively build the left and right subtrees root->left = buildBalancedTree(nodes, start, mid - 1); root->right = buildBalancedTree(nodes, mid + 1, end); return root; } // Function to balance a BST Node* balanceBST(Node* root) { vector<int> nodes; // Store the nodes in sorted order storeInorder(root, nodes); // Build the balanced tree from the sorted nodes return buildBalancedTree(nodes, 0, nodes.size() - 1); } //Driver Code Starts // Print tree as level order void printLevelOrder(Node *root) { if (root == nullptr) { cout << "N "; return; } queue<Node *> qq; qq.push(root); int nonNull = 1; while (!qq.empty() && nonNull > 0) { Node *curr = qq.front(); qq.pop(); if (curr == nullptr) { cout << "N "; continue; } nonNull--; cout << (curr->data) << " "; qq.push(curr->left); qq.push(curr->right); if (curr->left) nonNull++; if (curr->right) nonNull++; } } int main() { // Constructing an unbalanced BST // 4 // / \ // 3 5 // / \ // 2 6 // / \ // 1 7 Node* root = new Node(4); root->left = new Node(3); root->left->left = new Node(2); root->left->left->left = new Node(1); root->right = new Node(5); root->right->right = new Node(6); root->right->right->right = new Node(7); Node* balancedRoot = balanceBST(root); printLevelOrder(balancedRoot); return 0; } //Driver Code Ends Java //Driver Code Starts // Java program to convert a left unbalanced BST to // a balanced BST import java.util.*; class Node { int data; Node left, right; Node(int value) { data = value; left = null; right = null; } } class GFG { //Driver Code Ends // Inorder traversal to store elements of the // tree in sorted order static void storeInorder(Node root, ArrayList<Integer> nodes) { if (root == null) return; // Traverse the left subtree storeInorder(root.left, nodes); // Store the node data nodes.add(root.data); // Traverse the right subtree storeInorder(root.right, nodes); } // Function to build a balanced BST from a sorted array static Node buildBalancedTree(ArrayList<Integer> nodes, int start, int end) { // Base case if (start > end) return null; // Get the middle element and make it the root int mid = (start + end) / 2; Node root = new Node(nodes.get(mid)); // Recursively build the left and right subtrees root.left = buildBalancedTree(nodes, start, mid - 1); root.right = buildBalancedTree(nodes, mid + 1, end); return root; } // Function to balance a BST static Node balanceBST(Node root) { ArrayList<Integer> nodes = new ArrayList<>(); // Store the nodes in sorted order storeInorder(root, nodes); // Build the balanced tree from the sorted nodes return buildBalancedTree(nodes, 0, nodes.size() - 1); } //Driver Code Starts // Print tree as level order static void printLevelOrder(Node root) { if (root == null) { System.out.print("N "); return; } Queue<Node> qq = new LinkedList<>(); qq.add(root); int nonNull = 1; while (!qq.isEmpty() && nonNull > 0) { Node curr = qq.poll(); if (curr == null) { System.out.print("N "); continue; } nonNull--; System.out.print(curr.data + " "); qq.add(curr.left); qq.add(curr.right); if (curr.left != null) nonNull++; if (curr.right != null) nonNull++; } } public static void main(String[] args) { // Constructing an unbalanced BST // 4 // / \ // 3 5 // / \ // 2 6 // / \ // 1 7 Node root = new Node(4); root.left = new Node(3); root.left.left = new Node(2); root.left.left.left = new Node(1); root.right = new Node(5); root.right.right = new Node(6); root.right.right.right = new Node(7); Node balancedRoot = balanceBST(root); printLevelOrder(balancedRoot); } } //Driver Code Ends Python #Driver Code Starts # Python program to convert a left unbalanced BST to # a balanced BST class Node: def __init__(self, value): self.data = value self.left = None self.right = None #Driver Code Ends # Inorder traversal to store elements of the # tree in sorted order def storeInorder(root, nodes): if root is None: return # Traverse the left subtree storeInorder(root.left, nodes) # Store the node data nodes.append(root.data) # Traverse the right subtree storeInorder(root.right, nodes) # Function to build a balanced BST from a sorted array def buildBalancedTree(nodes, start, end): # Base case if start > end: return None # Get the middle element and make it the root mid = (start + end) // 2 root = Node(nodes[mid]) # Recursively build the left and right subtrees root.left = buildBalancedTree(nodes, start, mid - 1) root.right = buildBalancedTree(nodes, mid + 1, end) return root # Function to balance a BST def balanceBST(root): nodes = [] # Store the nodes in sorted order storeInorder(root, nodes) # Build the balanced tree from the sorted nodes return buildBalancedTree(nodes, 0, len(nodes) - 1) #Driver Code Starts # Print tree as level order from collections import deque def printLevelOrder(root): if root is None: print("N", end=" ") return queue = deque([root]) nonNull = 1 while queue and nonNull > 0: curr = queue.popleft() if curr is None: print("N", end=" ") continue nonNull -= 1 print(curr.data, end=" ") queue.append(curr.left) queue.append(curr.right) if curr.left: nonNull += 1 if curr.right: nonNull += 1 if __name__ == "__main__": root = Node(4) root.left = Node(3) root.left.left = Node(2) root.left.left.left = Node(1) root.right = Node(5) root.right.right = Node(6) root.right.right.right = Node(7) balancedRoot = balanceBST(root) printLevelOrder(balancedRoot) #Driver Code Ends C# //Driver Code Starts // C# program to convert a left unbalanced BST to // a balanced BST using System; using System.Collections.Generic; class Node { public int data; public Node left, right; public Node(int value) { data = value; left = null; right = null; } } class GFG { //Driver Code Ends // Inorder traversal to store elements of the // tree in sorted order static void storeInorder(Node root, List<int> nodes) { if (root == null) return; // Traverse the left subtree storeInorder(root.left, nodes); // Store the node data nodes.Add(root.data); // Traverse the right subtree storeInorder(root.right, nodes); } // Function to build a balanced BST from a sorted array static Node buildBalancedTree(List<int> nodes, int start, int end) { // Base case if (start > end) return null; // Get the middle element and make it the root int mid = (start + end) / 2; Node root = new Node(nodes[mid]); // Recursively build the left and right subtrees root.left = buildBalancedTree(nodes, start, mid - 1); root.right = buildBalancedTree(nodes, mid + 1, end); return root; } // Function to balance a BST static Node balanceBST(Node root) { List<int> nodes = new List<int>(); // Store the nodes in sorted order storeInorder(root, nodes); // Build the balanced tree from the sorted nodes return buildBalancedTree(nodes, 0, nodes.Count - 1); } //Driver Code Starts // Print tree as level order static void printLevelOrder(Node root) { if (root == null) { Console.Write("N "); return; } Queue<Node> qq = new Queue<Node>(); qq.Enqueue(root); int nonNull = 1; while (qq.Count > 0 && nonNull > 0) { Node curr = qq.Dequeue(); if (curr == null) { Console.Write("N "); continue; } nonNull--; Console.Write(curr.data + " "); qq.Enqueue(curr.left); qq.Enqueue(curr.right); if (curr.left != null) nonNull++; if (curr.right != null) nonNull++; } } static void Main() { // Constructing an unbalanced BST // 4 // / \ // 3 5 // / \ // 2 6 // / \ // 1 7 Node root = new Node(4); root.left = new Node(3); root.left.left = new Node(2); root.left.left.left = new Node(1); root.right = new Node(5); root.right.right = new Node(6); root.right.right.right = new Node(7); Node balancedRoot = balanceBST(root); printLevelOrder(balancedRoot); } } //Driver Code Ends JavaScript //Driver Code Starts // JavaScript program to convert a left unbalanced BST to // a balanced BST class Node { constructor(value) { this.data = value; this.left = null; this.right = null; } } //Driver Code Ends // Inorder traversal to store elements of the // tree in sorted order function storeInorder(root, nodes) { if (root === null) return; // Traverse the left subtree storeInorder(root.left, nodes); // Store the node data nodes.push(root.data); // Traverse the right subtree storeInorder(root.right, nodes); } // Function to build a balanced BST from a sorted array function buildBalancedTree(nodes, start, end) { // Base case if (start > end) return null; // Get the middle element and make it the root let mid = Math.floor((start + end) / 2); let root = new Node(nodes[mid]); // Recursively build the left and right subtrees root.left = buildBalancedTree(nodes, start, mid - 1); root.right = buildBalancedTree(nodes, mid + 1, end); return root; } // Function to balance a BST function balanceBST(root) { let nodes = []; // Store the nodes in sorted order storeInorder(root, nodes); // Build the balanced tree from the sorted nodes return buildBalancedTree(nodes, 0, nodes.length - 1); } //Driver Code Starts // Print tree as level order function printLevelOrder(root) { if (root === null) { console.log("N"); return; } let queue = []; queue.push(root); let nonNull = 1; while (queue.length > 0 && nonNull > 0) { let curr = queue.shift(); if (curr === null) { process.stdout.write("N "); continue; } nonNull--; process.stdout.write(curr.data + " "); queue.push(curr.left); queue.push(curr.right); if (curr.left) nonNull++; if (curr.right) nonNull++; } } // Driver Code // Constructing an unbalanced BST // 4 // / \ // 3 5 // / \ // 2 6 // / \ // 1 7 let root = new Node(4); root.left = new Node(3); root.left.left = new Node(2); root.left.left.left = new Node(1); root.right = new Node(5); root.right.right = new Node(6); root.right.right.right = new Node(7); let balancedRoot = balanceBST(root); printLevelOrder(balancedRoot); //Driver Code Ends Output4 2 6 1 3 5 7 Time Complexity: O(n), as we are just traversing the tree twice. Once in inorder traversal and then in construction of the balanced tree.Auxiliary space: O(n), as extra space is used to store the nodes of the inorder traversal in the vector. Also the extra space taken by recursion call stack is O(h) where h is the height of the tree. Balance a Binary Search Tree Visit Course Comment More infoAdvertise with us Next Article Analysis of Algorithms K kartik Follow Improve Article Tags : Tree Binary Search Tree DSA Practice Tags : Binary Search TreeTree Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. 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