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Convert a Binary Tree to BST by left shifting digits of node values
  • Difficulty Level : Easy
  • Last Updated : 28 Jan, 2021

Given a Binary Tree of positive integers. The task is to convert it to a BST using left shift operations on the digits of the nodes. If it is not possible to convert the binary tree to BST then print -1.

Examples:

Input:                      443 
                              /        \  
                          132      543 
                                    /      \ 
                                343     237

Output:                   344 
                               /        \ 
                         132       435 
                                   /        \ 
                               433       723

Explanation:   443 → left shift two times → 344 
                         132 → zero shift operations → 132 
                         543 → left shift one time → 435 
                         343 → left shift one time → 433 
                         237 → left shift two times → 723



Input:                    43 
                         /        \ 
                      56         45

Output: -1

Approach: The idea is to perform Inorder traversal on the Binary Tree in increasing order because in-order traversal of a BST always generates sorted sequence. Follow the steps below to solve the problem:

  • Initialize a variable, say prev = -1, to keep track of the value of the previous node.
  • Then, traverse the tree using Inorder Traversal and try to convert every node to its lowest possible value greater than the previous value by left shifting its digits.
  • After performing these operations, check if the binary tree is a BST or not.
  • If found to be true, print the tree. Otherwise, print -1.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Structure of a Node
struct TreeNode
{
   int val;
   TreeNode *left, *right;
 
    TreeNode(int key)
    {
        val = key;
        left = NULL;
        right = NULL;
    }
};
 
// Function to check if
// the tree is BST or not
bool isBST(TreeNode *root, TreeNode *l = NULL, TreeNode *r = NULL)
{
 
    // Base condition
    if (!root)
        return true;
 
    // Check for left child
    if (l && root->val <= l->val)
        return false;
 
    // Check for right child
    if (r && root->val >= r->val)
        return false;
 
    // Check for left and right subtrees
    return isBST(root->left, l, root) and isBST(root->right, root, r);
}
 
// Function to convert a tree to BST
// by left shift of its digits
void ConvTree(TreeNode *root,int &prev)
{
 
    // If root is NULL
    if (!root)
        return;
 
    // Recursively modify the
    // left subtree
    ConvTree(root->left,prev);
 
    // Initialise optimal element
    // to the initial element
    int optEle = root->val;
 
    // Convert it into a string
    string strEle = to_string(root->val);
 
    // For checking all the
    // left shift operations
    bool flag = true;
    for (int idx = 0; idx < strEle.size(); idx++)
    {
 
        // Perform left shift
        int shiftedNum = stoi(strEle.substr(idx) + strEle.substr(0, idx));
 
        // If the number after left shift
        // is the minimum and greater than
        // its previous element
 
        // first value >= prev
 
        // used to setup initialize optEle
        // cout<<prev<<endl;
        if (shiftedNum >= prev and flag)
        {
            optEle = shiftedNum;
            flag = false;
          }
 
        if (shiftedNum >= prev)
            optEle = min(optEle, shiftedNum);
      }
    root->val = optEle;
    prev = root->val;
   
    // Recursively solve for right
    // subtree
    ConvTree(root->right,prev);
}
 
// Function to print level
// order traversal of a tree
void levelOrder(TreeNode *root)
{
    queue<TreeNode*> que;
    que.push(root);
    while(true)
    {
        int length = que.size();
        if (length == 0)
            break;
        while (length)
        {
            auto temp = que.front();
            que.pop();
            cout << temp->val << " ";
            if (temp->left)
                que.push(temp->left);
            if (temp->right)
                que.push(temp->right);
            length -= 1;
          }
          cout << endl;
        }
   
        // new line
        cout << endl;
}
 
// Driver Code
int main()
{
 
  TreeNode *root = new TreeNode(443);
  root->left = new TreeNode(132);
  root->right = new TreeNode(543);
  root->right->left = new TreeNode(343);
  root->right->right = new TreeNode(237);
 
  // Converts the tree to BST
  int prev = -1;
  ConvTree(root, prev);
 
  // If tree is converted to a BST
  if (isBST(root, NULL, NULL))
  {
      // Print its level order traversal
      levelOrder(root);
    }
  else
  {
     
      // not possible
      cout << (-1);
    }
}
 
// This code is contributed by mohit kumar 29

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Java

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// Java program for the above approach
import java.util.*;
 
class GFG{
     
static int prev;
 
// Structure of a Node
static class TreeNode
{
    int val;
    TreeNode left, right;
     
    TreeNode(int key)
    {
        val = key;
        left = null;
        right = null;
    }
};
 
// Function to check if
// the tree is BST or not
static boolean isBST(TreeNode root, TreeNode l,
                     TreeNode r)
{
     
    // Base condition
    if (root == null)
        return true;
 
    // Check for left child
    if (l != null && root.val <= l.val)
        return false;
 
    // Check for right child
    if (r != null && root.val >= r.val)
        return false;
 
    // Check for left and right subtrees
    return isBST(root.left, l, root) &
           isBST(root.right, root, r);
}
 
// Function to convert a tree to BST
// by left shift of its digits
static void ConvTree(TreeNode root)
{
     
    // If root is null
    if (root == null)
        return;
         
    // Recursively modify the
    // left subtree
    ConvTree(root.left);
 
    // Initialise optimal element
    // to the initial element
    int optEle = root.val;
 
    // Convert it into a String
    String strEle = String.valueOf(root.val);
 
    // For checking all the
    // left shift operations
    boolean flag = true;
     
    for(int idx = 0; idx < strEle.length(); idx++)
    {
         
        // Perform left shift
        int shiftedNum = Integer.parseInt(
            strEle.substring(idx) +
            strEle.substring(0, idx));
 
        // If the number after left shift
        // is the minimum and greater than
        // its previous element
 
        // first value >= prev
 
        // used to setup initialize optEle
        // System.out.print(prev<<endl;
        if (shiftedNum >= prev && flag)
        {
            optEle = shiftedNum;
            flag = false;
        }
         
        if (shiftedNum >= prev)
            optEle = Math.min(optEle, shiftedNum);
    }
    root.val = optEle;
    prev = root.val;
   
    // Recursively solve for right
    // subtree
    ConvTree(root.right);
}
 
// Function to print level
// order traversal of a tree
static void levelOrder(TreeNode root)
{
    Queue<TreeNode> que = new LinkedList<GFG.TreeNode>();
    que.add(root);
     
    while(true)
    {
        int length = que.size();
         
        if (length == 0)
            break;
             
        while (length > 0)
        {
            TreeNode temp = que.peek();
            que.remove();
            System.out.print(temp.val + " ");
             
            if (temp.left != null)
                que.add(temp.left);
            if (temp.right != null)
                que.add(temp.right);
                 
            length -= 1;
        }
        System.out.println();
    }
     
    // New line
    System.out.println();
}
 
// Driver Code
public static void main(String[] args)
{
    TreeNode root = new TreeNode(443);
    root.left = new TreeNode(132);
    root.right = new TreeNode(543);
    root.right.left = new TreeNode(343);
    root.right.right = new TreeNode(237);
     
    // Converts the tree to BST
    prev = -1;
    ConvTree(root);
     
    // If tree is converted to a BST
    if (isBST(root, null, null))
    {
         
        // Print its level order traversal
        levelOrder(root);
    }
    else
    {
         
        // not possible
        System.out.print(-1);
    }
}
}
 
// This code is contributed by shikhasingrajput

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Python3

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# Python3 program for the above approach
 
# Structure of a Node
class TreeNode:
    def __init__(self, key):
        self.val = key
        self.left = None
        self.right = None
 
# Function to check if
# the tree is BST or not
def isBST(root, l = None, r = None):
 
    # Base condition
    if not root:
        return True
 
    # Check for left child
    if l and root.val <= l.val:
        return False
 
    # Check for right child
    if r and root.val >= r.val:
        return False
 
    # Check for left and right subtrees
    return isBST(root.left, l, root) and isBST(root.right, root, r)
 
# Function to convert a tree to BST
# by left shift of its digits
def ConvTree(root):
    global prev
 
    # If root is NULL
    if not root:
        return
 
    # Recursively modify the
    # left subtree
    ConvTree(root.left)
 
    # Initialise optimal element
    # to the initial element
    optEle = root.val
 
    # Convert it into a string
    strEle = str(root.val)
 
    # For checking all the
    # left shift operations
    flag = True
 
    for idx in range(len(strEle)):
 
        # Perform left shift
        shiftedNum = int(strEle[idx:] + strEle[:idx])
 
        # If the number after left shift
        # is the minimum and greater than
        # its previous element
 
        # first value >= prev
 
        # used to setup initialize optEle
        if shiftedNum >= prev and flag:
            optEle = shiftedNum
            flag = False
 
        if shiftedNum >= prev:
            optEle = min(optEle, shiftedNum)
 
    root.val = optEle
    prev = root.val
    # Recursively solve for right
    # subtree
    ConvTree(root.right)
 
# Function to print level
# order traversal of a tree
def levelOrder(root):
    que = [root]
    while True:
        length = len(que)
        if not length:
            break
 
        while length:
            temp = que.pop(0)
            print(temp.val, end =' ')
            if temp.left:
                que.append(temp.left)
            if temp.right:
                que.append(temp.right)
            length -= 1
        # new line
        print()
 
# Driver Code
 
root = TreeNode(443)
root.left = TreeNode(132)
root.right = TreeNode(543)
root.right.left = TreeNode(343)
root.right.right = TreeNode(237)
 
# Initialize to
# previous element
prev = -1
 
# Converts the tree to BST
ConvTree(root)
 
# If tree is converted to a BST
if isBST(root, None, None):
 
    # Print its level order traversal
    levelOrder(root)
else:
    # not possible
    print(-1)

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C#

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// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
     
static int prev;
 
// Structure of a Node
class TreeNode
{
    public int val;
    public TreeNode left, right; 
    public TreeNode(int key)
    {
        val = key;
        left = null;
        right = null;
    }
};
 
// Function to check if
// the tree is BST or not
static bool isBST(TreeNode root, TreeNode l,
                     TreeNode r)
{
     
    // Base condition
    if (root == null)
        return true;
 
    // Check for left child
    if (l != null && root.val <= l.val)
        return false;
 
    // Check for right child
    if (r != null && root.val >= r.val)
        return false;
 
    // Check for left and right subtrees
    return isBST(root.left, l, root) &
           isBST(root.right, root, r);
}
 
// Function to convert a tree to BST
// by left shift of its digits
static void ConvTree(TreeNode root)
{
     
    // If root is null
    if (root == null)
        return;
         
    // Recursively modify the
    // left subtree
    ConvTree(root.left);
 
    // Initialise optimal element
    // to the initial element
    int optEle = root.val;
 
    // Convert it into a String
    String strEle = String.Join("", root.val);
 
    // For checking all the
    // left shift operations
    bool flag = true;
     
    for(int idx = 0; idx < strEle.Length; idx++)
    {
         
        // Perform left shift
        int shiftedNum = Int32.Parse(
            strEle.Substring(idx) +
            strEle.Substring(0, idx));
 
        // If the number after left shift
        // is the minimum and greater than
        // its previous element
 
        // first value >= prev
 
        // used to setup initialize optEle
        // Console.Write(prev<<endl;
        if (shiftedNum >= prev && flag)
        {
            optEle = shiftedNum;
            flag = false;
        }
         
        if (shiftedNum >= prev)
            optEle = Math.Min(optEle, shiftedNum);
    }
    root.val = optEle;
    prev = root.val;
   
    // Recursively solve for right
    // subtree
    ConvTree(root.right);
}
 
// Function to print level
// order traversal of a tree
static void levelOrder(TreeNode root)
{
    Queue<TreeNode> que = new Queue<GFG.TreeNode>();
    que.Enqueue(root);
     
    while(true)
    {
        int length = que.Count;
        if (length == 0)
            break;     
        while (length > 0)
        {
            TreeNode temp = que.Peek();
            que.Dequeue();
            Console.Write(temp.val + " ");      
            if (temp.left != null)
                que.Enqueue(temp.left);
            if (temp.right != null)
                que.Enqueue(temp.right);       
            length -= 1;
        }
        Console.WriteLine();
    }
     
    // New line
    Console.WriteLine();
}
 
// Driver Code
public static void Main(String[] args)
{
    TreeNode root = new TreeNode(443);
    root.left = new TreeNode(132);
    root.right = new TreeNode(543);
    root.right.left = new TreeNode(343);
    root.right.right = new TreeNode(237);
     
    // Converts the tree to BST
    prev = -1;
    ConvTree(root);
     
    // If tree is converted to a BST
    if (isBST(root, null, null))
    {
         
        // Print its level order traversal
        levelOrder(root);
    }
    else
    {
         
        // not possible
        Console.Write(-1);
    }
}
}
  
// This code is contributed by shikhasingrajput

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Output: 

344 
132 435 
433 723

 

Time Complexity: O(N)
Auxiliary Space: O(N)

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