Convert a normal BST to Balanced BST - GeeksforGeeks

23 Aug, 2021
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.  
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.
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++
Java
Python3
C#
Javascript
// 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 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);         // 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 program to convert a left# unbalanced BST to a balanced BSTimport sysimport math # A binary tree node has data, pointer to left child# and a pointer to right childclass 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 treedef 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
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
<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 
 
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