Convert a normal BST to Balanced BST

2.7

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

An Efficient Solution can construct balanced BST in O(n) time with minimum possible height. Below are steps.

  1. 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.
  2. 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 C++ 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 BST
Node* buildTree(Node* root)
{
    // Store nodes of given BST in sorted order
    vector<Node *> nodes;
    storeBSTNodes(root, nodes);

    // Constucts BST from nodes[]
    int n = nodes.size();
    return buildTreeUtil(nodes, 0, n-1);
}

// Utility function to create a new node
Node* 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 program
int 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 BST

import 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);

        // Constucts 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)

Output :

Preorder traversal of balanced BST is : 
7 5 6 8 10 

This article is contributed Aditya Goel. If you likeGeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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