Binary Tree Upside Down

Binary Tree Upside Down

What is this problem about?

The Binary Tree Upside Down interview question is a unique structural transformation problem. Given a binary tree where every right node either is a leaf with a sibling (a left node) or is empty, you need to flip it upside down. After the flip, the original left child becomes the new root, the original root becomes the new right child, and the original right sibling becomes the new left child.

Why is this asked in interviews?

LinkedIn frequently uses this problem to test a candidate's ability to manipulate pointers and visualize complex tree rotations. Unlike standard traversals, this problem requires a deep understanding of how to reassign pointers in a specific order without losing the reference to the rest of the tree. It’s essentially a variation of the "Reverse a Linked List" problem but applied to a binary tree structure.

Algorithmic pattern used

You can solve this using either Recursion or an Iterative approach. The pattern is similar to reversing a linked list. In the recursive approach, you go down to the leftmost node (which will become the new root) and then, as the recursion unwinds, you reassign the pointers of the current node's children to point back to the current node and its right sibling.

Example explanation

Original tree: 1 / 2 3 / 4 5

Flipped result: 4 / 5 2 / 3 1

1. The leftmost node (4) becomes the new root.
2. 4's left child becomes 5 (its original right sibling).
3. 4's right child becomes 2 (its original parent).
4. Similarly, 2's new left child is 3 and its new right child is 1.

Common mistakes candidates make

The most common mistake is losing the reference to a node before reassigning its pointers, leading to a disconnected tree. Another error is not handling the base case correctly—only the leftmost leaf should become the new root. Candidates also often struggle with the "upside down" visualization, as it’s not a standard rotation.

Interview preparation tip

Draw the tree before and after the transformation. Label each node and trace where each pointer (left and right) goes. This visualization is key to getting the pointer assignment logic right in your code.