Given a binary search tree, the task is to flatten it to a sorted list in decreasing order. Precisely, the value of each node must be greater 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.
Input: 5 / \ 3 7 / \ / \ 2 4 6 8 Output: 8 7 6 5 4 3 2 Input: 1 \ 2 \ 3 \ 4 \ 5 Output: 5 4 3 2 1
Approach: A simple approach will be to recreate the BST from its ‘reverse 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 the reverse in-order traversal of a binary tree as follows:
- Create a dummy node.
- Create variable called ‘prev’ and make it point to dummy node.
- Perform reverse 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 the worst case as in-order traversal takes O(H) extra space.
Below is the implementation of the above approach:
8 7 6 5 4 3 2
- Flatten BST to sorted list | Increasing order
- Flatten Binary Tree in order of Level Order Traversal
- Flatten binary tree in order of post-order traversal
- Flatten Binary Tree in order of Zig Zag traversal
- Flatten a binary tree into linked list | Set-2
- Flatten a binary tree into linked list | Set-3
- Flatten a binary tree into linked list
- Print array elements in alternatively increasing and decreasing order
- Sorted order printing of a given array that represents a BST
- Print Binary Tree levels in sorted order | Set 2 (Using set)
- Print Binary Tree levels in sorted order
- Print Binary Tree levels in sorted order | Set 3 (Tree given as array)
- Sorted Linked List to Balanced BST
- Print the last k nodes of the linked list in reverse order | Iterative Approaches
- Count triplets in a sorted doubly linked list whose product is equal to a given value x
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