Flatten BST to sorted list | Increasing order

17 Dec, 2021
Given a binary search tree, the task is to flatten it to a sorted list. Precisely, the value of each node must be lesser than the values of all the nodes at its right, and its left node must be NULL after flattening. We must do it in O(H) extra space where ‘H’ is the height of BST.
Examples: 
Input: 
          5 
        /   \ 
       3     7 
      / \   / \ 
     2   4 6   8
Output: 2 3 4 5 6 7 8
Input:
      1
       \
        2
         \
          3
           \
            4
             \
              5
Output: 1 2 3 4 5
Approach: A simple approach will be to recreate the BST from its in-order traversal. This will take O(N) extra space where N is the number of nodes in BST. 
To improve upon that, we will simulate in-order traversal of a binary tree as follows:  
Create a dummy node.Create a variable called ‘prev’ and make it point to the dummy node.Perform in-order traversal and at each step. Set prev -> right = currSet prev -> left = NULLSet prev = curr
Create a dummy node.
Create a variable called ‘prev’ and make it point to the dummy node.
Perform in-order traversal and at each step. Set prev -> right = currSet prev -> left = NULLSet prev = curr
Set prev -> right = curr
Set prev -> left = NULL
Set prev = curr
This will improve the space complexity to O(H) in worst case as in-order traversal takes O(H) extra space.
Below is the implementation of the above approach: 
C++
Java
C#
Javascript
Python3
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std; // Node of the binary treestruct node {    int data;    node* left;    node* right;    node(int data)    {        this->data = data;        left = NULL;        right = NULL;    }}; // Function to print flattened// binary Treevoid print(node* parent){    node* curr = parent;    while (curr != NULL)        cout << curr->data << " ", curr = curr->right;} // Function to perform in-order traversal// recursivelyvoid inorder(node* curr, node*& prev){    // Base case    if (curr == NULL)        return;    inorder(curr->left, prev);    prev->left = NULL;    prev->right = curr;    prev = curr;    inorder(curr->right, prev);} // Function to flatten binary tree using// level order traversalnode* flatten(node* parent){    // Dummy node    node* dummy = new node(-1);     // Pointer to previous element    node* prev = dummy;     // Calling in-order traversal    inorder(parent, prev);     prev->left = NULL;    prev->right = NULL;    node* ret = dummy->right;     // Delete dummy node    delete dummy;    return ret;} // Driver codeint main(){    node* root = new node(5);    root->left = new node(3);    root->right = new node(7);    root->left->left = new node(2);    root->left->right = new node(4);    root->right->left = new node(6);    root->right->right = new node(8);     // Calling required function    print(flatten(root));     return 0;}
// Java implementation of the// above approachimport java.util.*;class GFG{  // Node of the binary treestatic class node{  int data;  node left;  node right;      node(int data)  {    this.data = data;    left = null;    right = null;  }};  // Function to print flattened// binary treestatic void print(node parent){  node curr = parent;  while (curr != null)  {    System.out.print(curr.data + " ");    curr = curr.right;  }}  static  node prev;    // Function to perform// in-order traversalstatic void Inorder(node curr){  // Base case  if (curr == null)    return;  Inorder(curr.left);  prev.left = null;  prev.right = curr;  prev = curr;  Inorder(curr.right);}  // Function to flatten binary// tree using level order// traversalstatic node flatten(node parent){  // Dummy node  node dummy = new node(-1);    // Pointer to previous  // element  prev = dummy;    // Calling in-order  // traversal  Inorder(parent);    prev.left = null;  prev.right = null;  node ret = dummy.right;    // Delete dummy node  //delete dummy;  return ret;}  // Driver codepublic static void main(String[] args){  node root = new node(5);  root.left = new node(3);  root.right = new node(7);  root.left.left = new node(2);  root.left.right = new node(4);  root.right.left = new node(6);  root.right.right = new node(8);    // Calling required function  print(flatten(root));}}  // This code is contributed by Debojyoti Mandal
// C# implementation of the// above approachusing System;public class Program{  // Node of the binary treepublic class node{  public int data;  public node left;  public node right;      public node(int data)  {    this.data = data;    left = null;    right = null;  }};  // Function to print flattened// binary treestatic void print(node parent){  node curr = parent;  while (curr != null)  {    Console.Write(curr.data + " ");    curr = curr.right;  }}  static  node prev;    // Function to perform// in-order traversalstatic void Inorder(node curr){  // Base case  if (curr == null)    return;  Inorder(curr.left);  prev.left = null;  prev.right = curr;  prev = curr;  Inorder(curr.right);}  // Function to flatten binary// tree using level order// traversalstatic node flatten(node parent){  // Dummy node  node dummy = new node(-1);    // Pointer to previous  // element  prev = dummy;    // Calling in-order  // traversal  Inorder(parent);    prev.left = null;  prev.right = null;  node ret = dummy.right;    // Delete dummy node  //delete dummy;  return ret;}  // Driver codepublic static void Main(string[] args){  node root = new node(5);  root.left = new node(3);  root.right = new node(7);  root.left.left = new node(2);  root.left.right = new node(4);  root.right.left = new node(6);  root.right.right = new node(8);    // Calling required function  print(flatten(root));}} // This code is contributed by rrrtnx.
<script> // Javascript implementation of the approach // Node of the binary treeclass node{    constructor(data)    {        this.left = null;        this.right = null;        this.data = data;    }} let prev; // Function to print flattened// binary Treefunction print(parent){    let curr = parent;    while (curr != null)    {        document.write(curr.data + " ");        curr = curr.right;    }} // Function to perform in-order traversal// recursivelyfunction inorder(curr){         // Base case    if (curr == null)        return;             inorder(curr.left);    prev.left = null;    prev.right = curr;    prev = curr;    inorder(curr.right);} // Function to flatten binary tree using// level order traversalfunction flatten(parent){         // Dummy node    let dummy = new node(-1);     // Pointer to previous element    prev = dummy;     // Calling in-order traversal    inorder(parent);     prev.left = null;    prev.right = null;    let ret = dummy.right;     // Delete dummy node    return ret;} // Driver codelet root = new node(5);root.left = new node(3);root.right = new node(7);root.left.left = new node(2);root.left.right = new node(4);root.right.left = new node(6);root.right.right = new node(8); // Calling required functionprint(flatten(root)); // This code is contributed by divyeshrabadiya07 </script>
# Python3 implementation of the approach global prev# Node of the binary treeclass node :    def __init__(self, data):        self.data = data        self.left = None        self.right = None       # Function to print flattened# binary Treedef printTree(parent):    curr = parent    while (curr != None):        print(curr.data,end=' ')        curr = curr.right  # Function to perform in-order traversal# recursivelydef inorder(curr):    global prev    # Base case    if (curr == None):        return    inorder(curr.left)    prev.left = None    prev.right = curr    prev = curr    inorder(curr.right)  # Function to flatten binary tree using# level order traversaldef flatten(parent):    global prev    # Dummy node    dummy = node(-1)     # Pointer to previous element    prev = dummy     # Calling in-order traversal    inorder(parent)     prev.left = None    prev.right = None    ret = dummy.right     # Delete dummy node    return ret  # Driver codeif __name__ == '__main__':    root = node(5)    root.left = node(3)    root.right = node(7)    root.left.left = node(2)    root.left.right = node(4)    root.right.left = node(6)    root.right.right = node(8)     # Calling required function    printTree(flatten(root))
2 3 4 5 6 7 8
 
Time Complexity: O(N)Auxiliary Space: O(N)
divyeshrabadiya07
mahinsagotra18
pankajsharmagfg
jyoti369
rrrtnx
amartyaghoshgfg
Algorithms
Binary Search Tree
Tree
Binary Search Tree
Tree
Algorithms
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