Reverse Odd Levels of Binary Tree

Reverse Odd Levels of Binary Tree

What is this problem about?

The Reverse Odd Levels of Binary Tree interview question asks you to reverse the node values at all odd-depth levels of a perfect binary tree (where depth 0 is the root, depth 1 is odd, depth 2 is even, etc.). The tree structure remains unchanged — only the values at odd levels are reversed. Return the root of the modified tree.

Why is this asked in interviews?

This problem is asked at J.P. Morgan, Microsoft, Amazon, Google, and Jpmorgan because it tests multi-level tree traversal with level-based processing. It combines BFS (for level-order traversal) or DFS (for symmetric pair swapping) with the ability to selectively modify values at specific levels. It evaluates understanding of perfect binary tree properties and level parity.

Algorithmic pattern used

Two clean approaches: BFS and DFS. BFS approach: perform level-order traversal, collecting all nodes at each level into a list. For odd levels, reverse the values (not the nodes themselves — swap values between corresponding positions). DFS approach: use a recursive function that takes two symmetric nodes and their depth. If depth is odd, swap their values. Then recurse on (left.left, right.right) and (left.right, right.left) — the symmetric pairs at the next level.

Example explanation

Tree (perfect binary tree):

        2
      /   \
     3     5
    / \   / \
   8   13 21 34

Level 1 (odd): values [3, 5] → reverse → [5, 3]. Swap node values. Level 3 (odd): values [8, 13, 21, 34] → reverse → [34, 21, 13, 8].

Result:

        2
      /   \
     5     3
    / \   / \
  34  21 13  8

Common mistakes candidates make

• Reversing node pointers instead of node values — the problem only requires value swaps.
• Using BFS but accessing the wrong level index (off-by-one on level parity check).
• In the DFS approach, not pairing the correct symmetric nodes — (left.left, right.right) and (left.right, right.left) must be processed correctly.
• Not recognizing this problem requires a perfect binary tree — the algorithm assumes every level is fully populated.

Interview preparation tip

For the Reverse Odd Levels of Binary Tree coding problem, the BFS and DFS binary tree interview pattern both apply cleanly. The DFS approach is elegant: it treats the tree symmetrically by always passing pairs of mirror nodes. Practice writing the DFS version with the four recursive calls for mirror pairs. Google interviewers appreciate the DFS approach for its elegance and O(n) time with O(log n) recursion depth (since the tree is perfect and balanced).