Given a BST (Binary Search Tree) that may be unbalanced, convert it into a balanced BST that has minimum possible height.
Examples :
Input:
30
/
20
/
10
Output:
20
/ \
10 30
Input:
4
/
3
/
2
/
1
Output:
3 3 2
/ \ / \ / \
1 4 OR 2 4 OR 1 3 OR ..
\ / \
2 1 4
Input:
4
/ \
3 5
/ \
2 6
/ \
1 7
Output:
4
/ \
2 6
/ \ / \
1 3 5 7
A Simple Solution is to traverse nodes in Inorder and one by one insert into a self-balancing BST like AVL tree. Time complexity of this solution is O(n Log n) and this solution doesn’t guarantee the minimum possible height as in the worst case the height of the AVL tree can be 1.44*log2n.
An Efficient Solution can be to construct a balanced BST in O(n) time with minimum possible height. Below are steps.
- Traverse given BST in inorder and store result in an array. This step takes O(n) time. 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 here. This step also takes O(n) time as we traverse every element exactly once and processing an element takes O(1) time.
Below is the implementation of above steps.
C++
// C++ program to convert a left unbalanced BST to// a balanced BST#include <bits/stdc++.h>using namespace std;struct Node{ int data; Node* left, *right;};/* This function traverse the skewed binary tree and stores its nodes pointers in vector nodes[] */void storeBSTNodes(Node* root, vector<Node*> &nodes){ // Base case if (root==NULL) return; // Store nodes in Inorder (which is sorted // order for BST) storeBSTNodes(root->left, nodes); nodes.push_back(root); storeBSTNodes(root->right, nodes);}/* Recursive function to construct binary tree */Node* buildTreeUtil(vector<Node*> &nodes, int start, int end){ // base case if (start > end) return NULL; /* Get the middle element and make it root */ int mid = (start + end)/2; Node *root = nodes[mid]; /* Using index in Inorder traversal, construct left and right subtress */ root->left = buildTreeUtil(nodes, start, mid-1); root->right = buildTreeUtil(nodes, mid+1, end); return root;}// This functions converts an unbalanced BST to// a balanced BSTNode* buildTree(Node* root){ // Store nodes of given BST in sorted order vector<Node *> nodes; storeBSTNodes(root, nodes); // Constructs BST from nodes[] int n = nodes.size(); return buildTreeUtil(nodes, 0, n-1);}// Utility function to create a new nodeNode* newNode(int data){ Node* node = new Node; node->data = data; node->left = node->right = NULL; return (node);}/* Function to do preorder traversal of tree */void preOrder(Node* node){ if (node == NULL) return; printf("%d ", node->data); preOrder(node->left); preOrder(node->right);}// Driver programint main(){ /* Constructed skewed binary tree is 10 / 8 / 7 / 6 / 5 */ Node* root = newNode(10); root->left = newNode(8); root->left->left = newNode(7); root->left->left->left = newNode(6); root->left->left->left->left = newNode(5); root = buildTree(root); printf("Preorder traversal of balanced " "BST is : \n"); preOrder(root); return 0;} |
Java
// Java program to convert a left unbalanced BST to a balanced BSTimport java.util.*;/* A binary tree node has data, pointer to left child and a pointer to right child */class Node { int data; Node left, right; public Node(int data) { this.data = data; left = right = null; }}class BinaryTree { Node root; /* This function traverse the skewed binary tree and stores its nodes pointers in vector nodes[] */ void storeBSTNodes(Node root, Vector<Node> nodes) { // Base case if (root == null) return; // Store nodes in Inorder (which is sorted // order for BST) storeBSTNodes(root.left, nodes); nodes.add(root); storeBSTNodes(root.right, nodes); } /* Recursive function to construct binary tree */ Node buildTreeUtil(Vector<Node> nodes, int start, int end) { // base case if (start > end) return null; /* Get the middle element and make it root */ int mid = (start + end) / 2; Node node = nodes.get(mid); /* Using index in Inorder traversal, construct left and right subtress */ node.left = buildTreeUtil(nodes, start, mid - 1); node.right = buildTreeUtil(nodes, mid + 1, end); return node; } // This functions converts an unbalanced BST to // a balanced BST Node buildTree(Node root) { // Store nodes of given BST in sorted order Vector<Node> nodes = new Vector<Node>(); storeBSTNodes(root, nodes); // Constructs BST from nodes[] int n = nodes.size(); return buildTreeUtil(nodes, 0, n - 1); } /* Function to do preorder traversal of tree */ void preOrder(Node node) { if (node == null) return; System.out.print(node.data + " "); preOrder(node.left); preOrder(node.right); } // Driver program to test the above functions public static void main(String[] args) { /* Constructed skewed binary tree is 10 / 8 / 7 / 6 / 5 */ BinaryTree tree = new BinaryTree(); tree.root = new Node(10); tree.root.left = new Node(8); tree.root.left.left = new Node(7); tree.root.left.left.left = new Node(6); tree.root.left.left.left.left = new Node(5); tree.root = tree.buildTree(tree.root); System.out.println("Preorder traversal of balanced BST is :"); tree.preOrder(tree.root); }}// This code has been contributed by Mayank Jaiswal(mayank_24) |
Python3
# Python3 program to convert a left # unbalanced BST to a balanced BST import sysimport math# A binary tree node has data, pointer to left child # and a pointer to right child class Node: def __init__(self,data): self.data=data self.left=None self.right=None# This function traverse the skewed binary tree and # stores its nodes pointers in vector nodes[]def storeBSTNodes(root,nodes): # Base case if not root: return # Store nodes in Inorder (which is sorted # order for BST) storeBSTNodes(root.left,nodes) nodes.append(root) storeBSTNodes(root.right,nodes)# Recursive function to construct binary tree def buildTreeUtil(nodes,start,end): # base case if start>end: return None # Get the middle element and make it root mid=(start+end)//2 node=nodes[mid] # Using index in Inorder traversal, construct # left and right subtress node.left=buildTreeUtil(nodes,start,mid-1) node.right=buildTreeUtil(nodes,mid+1,end) return node# This functions converts an unbalanced BST to # a balanced BSTdef buildTree(root): # Store nodes of given BST in sorted order nodes=[] storeBSTNodes(root,nodes) # Constructs BST from nodes[] n=len(nodes) return buildTreeUtil(nodes,0,n-1)# Function to do preorder traversal of treedef preOrder(root): if not root: return print("{} ".format(root.data),end="") preOrder(root.left) preOrder(root.right)# Driver codeif __name__=='__main__': # Constructed skewed binary tree is # 10 # / # 8 # / # 7 # / # 6 # / # 5 root = Node(10) root.left = Node(8) root.left.left = Node(7) root.left.left.left = Node(6) root.left.left.left.left = Node(5) root = buildTree(root) print("Preorder traversal of balanced BST is :") preOrder(root) # This code has been contributed by Vikash Kumar 37 |
C#
using System;using System.Collections.Generic;// C# program to convert a left unbalanced BST to a balanced BST /* A binary tree node has data, pointer to left child and a pointer to right child */public class Node{ public int data; public Node left, right; public Node(int data) { this.data = data; left = right = null; }}public class BinaryTree{ public Node root; /* This function traverse the skewed binary tree and stores its nodes pointers in vector nodes[] */ public virtual void storeBSTNodes(Node root, List<Node> nodes) { // Base case if (root == null) { return; } // Store nodes in Inorder (which is sorted // order for BST) storeBSTNodes(root.left, nodes); nodes.Add(root); storeBSTNodes(root.right, nodes); } /* Recursive function to construct binary tree */ public virtual Node buildTreeUtil(List<Node> nodes, int start, int end) { // base case if (start > end) { return null; } /* Get the middle element and make it root */ int mid = (start + end) / 2; Node node = nodes[mid]; /* Using index in Inorder traversal, construct left and right subtress */ node.left = buildTreeUtil(nodes, start, mid - 1); node.right = buildTreeUtil(nodes, mid + 1, end); return node; } // This functions converts an unbalanced BST to // a balanced BST public virtual Node buildTree(Node root) { // Store nodes of given BST in sorted order List<Node> nodes = new List<Node>(); storeBSTNodes(root, nodes); // Constructs BST from nodes[] int n = nodes.Count; return buildTreeUtil(nodes, 0, n - 1); } /* Function to do preorder traversal of tree */ public virtual void preOrder(Node node) { if (node == null) { return; } Console.Write(node.data + " "); preOrder(node.left); preOrder(node.right); } // Driver program to test the above functions public static void Main(string[] args) { /* Constructed skewed binary tree is 10 / 8 / 7 / 6 / 5 */ BinaryTree tree = new BinaryTree(); tree.root = new Node(10); tree.root.left = new Node(8); tree.root.left.left = new Node(7); tree.root.left.left.left = new Node(6); tree.root.left.left.left.left = new Node(5); tree.root = tree.buildTree(tree.root); Console.WriteLine("Preorder traversal of balanced BST is :"); tree.preOrder(tree.root); }} // This code is contributed by Shrikant13 |
Javascript
<script> // JavaScript program to convert a left // unbalanced BST to a balanced BST class Node { constructor(data) { this.left = null; this.right = null; this.data = data; } } let root; /* This function traverse the skewed binary tree and stores its nodes pointers in vector nodes[] */ function storeBSTNodes(root, nodes) { // Base case if (root == null) return; // Store nodes in Inorder (which is sorted // order for BST) storeBSTNodes(root.left, nodes); nodes.push(root); storeBSTNodes(root.right, nodes); } /* Recursive function to construct binary tree */ function buildTreeUtil(nodes, start, end) { // base case if (start > end) return null; /* Get the middle element and make it root */ let mid = parseInt((start + end) / 2, 10); let node = nodes[mid]; /* Using index in Inorder traversal, construct left and right subtress */ node.left = buildTreeUtil(nodes, start, mid - 1); node.right = buildTreeUtil(nodes, mid + 1, end); return node; } // This functions converts an unbalanced BST to // a balanced BST function buildTree(root) { // Store nodes of given BST in sorted order let nodes = []; storeBSTNodes(root, nodes); // Constructs BST from nodes[] let n = nodes.length; return buildTreeUtil(nodes, 0, n - 1); } /* Function to do preorder traversal of tree */ function preOrder(node) { if (node == null) return; document.write(node.data + " "); preOrder(node.left); preOrder(node.right); } /* Constructed skewed binary tree is 10 / 8 / 7 / 6 / 5 */ root = new Node(10); root.left = new Node(8); root.left.left = new Node(7); root.left.left.left = new Node(6); root.left.left.left.left = new Node(5); root = buildTree(root); document.write("Preorder traversal of balanced BST is :" + "</br>"); preOrder(root); </script> |
Output
Preorder traversal of balanced BST is : 7 5 6 8 10
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), The 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.
This article is contributed Aditya Goel. If you likeneveropen and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
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