Related Articles

# Flatten BST to sorted list | Increasing order

• Difficulty Level : Medium
• Last Updated : 04 Aug, 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:

1. Create a dummy node.
2. Create a variable called ‘prev’ and make it point to the dummy node.
3. Perform in-order traversal and at each step.
• 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++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Node of the binary tree``struct` `node {``    ``int` `data;``    ``node* left;``    ``node* right;``    ``node(``int` `data)``    ``{``        ``this``->data = data;``        ``left = NULL;``        ``right = NULL;``    ``}``};` `// Function to print flattened``// binary Tree``void` `print(node* parent)``{``    ``node* curr = parent;``    ``while` `(curr != NULL)``        ``cout << curr->data << ``" "``, curr = curr->right;``}` `// Function to perform in-order traversal``// recursively``void` `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 traversal``node* 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 code``int` `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;``}`

## Javascript

 ``
Output:
`2 3 4 5 6 7 8`

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up