LeetCode #2415 — MEDIUM

Reverse Odd Levels of Binary Tree

Move from brute-force thinking to an efficient approach using tree strategy.

Solve on LeetCode

The Problem

Problem Statement

Given the root of a perfect binary tree, reverse the node values at each odd level of the tree.

For example, suppose the node values at level 3 are [2,1,3,4,7,11,29,18], then it should become [18,29,11,7,4,3,1,2].

Return the root of the reversed tree.

A binary tree is perfect if all parent nodes have two children and all leaves are on the same level.

The level of a node is the number of edges along the path between it and the root node.

Example 1:

Input: root = [2,3,5,8,13,21,34]
Output: [2,5,3,8,13,21,34]
Explanation: 
The tree has only one odd level.
The nodes at level 1 are 3, 5 respectively, which are reversed and become 5, 3.

Example 2:

Input: root = [7,13,11]
Output: [7,11,13]
Explanation: 
The nodes at level 1 are 13, 11, which are reversed and become 11, 13.

Example 3:

Input: root = [0,1,2,0,0,0,0,1,1,1,1,2,2,2,2]
Output: [0,2,1,0,0,0,0,2,2,2,2,1,1,1,1]
Explanation: 
The odd levels have non-zero values.
The nodes at level 1 were 1, 2, and are 2, 1 after the reversal.
The nodes at level 3 were 1, 1, 1, 1, 2, 2, 2, 2, and are 2, 2, 2, 2, 1, 1, 1, 1 after the reversal.

Constraints:

The number of nodes in the tree is in the range [1, 2¹⁴].
0 <= Node.val <= 10⁵
root is a perfect binary tree.

Step 01

Brute Force Baseline

Problem summary: Given the root of a perfect binary tree, reverse the node values at each odd level of the tree. For example, suppose the node values at level 3 are [2,1,3,4,7,11,29,18], then it should become [18,29,11,7,4,3,1,2]. Return the root of the reversed tree. A binary tree is perfect if all parent nodes have two children and all leaves are on the same level. The level of a node is the number of edges along the path between it and the root node.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Tree

Example 1

[2,3,5,8,13,21,34]

Example 2

[7,13,11]

Example 3

[0,1,2,0,0,0,0,1,1,1,1,2,2,2,2]

Core Insight

What unlocks the optimal approach

Try to solve recursively for each level independently.
While performing a depth-first search, pass the left and right nodes (which should be paired) to the next level. If the current level is odd, then reverse their values, or else recursively move to the next level.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2415: Reverse Odd Levels of Binary Tree
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode reverseOddLevels(TreeNode root) {
        Deque<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        for (int i = 0; !q.isEmpty(); ++i) {
            List<TreeNode> t = new ArrayList<>();
            for (int k = q.size(); k > 0; --k) {
                var node = q.poll();
                if (i % 2 == 1) {
                    t.add(node);
                }
                if (node.left != null) {
                    q.offer(node.left);
                    q.offer(node.right);
                }
            }
            for (int l = 0, r = t.size() - 1; l < r; ++l, --r) {
                var x = t.get(l).val;
                t.get(l).val = t.get(r).val;
                t.get(r).val = x;
            }
        }
        return root;
    }
}

// Accepted solution for LeetCode #2415: Reverse Odd Levels of Binary Tree
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func reverseOddLevels(root *TreeNode) *TreeNode {
	q := []*TreeNode{root}
	for i := 0; len(q) > 0; i++ {
		t := []*TreeNode{}
		for k := len(q); k > 0; k-- {
			node := q[0]
			q = q[1:]
			if i%2 == 1 {
				t = append(t, node)
			}
			if node.Left != nil {
				q = append(q, node.Left)
				q = append(q, node.Right)
			}
		}
		for l, r := 0, len(t)-1; l < r; l, r = l+1, r-1 {
			t[l].Val, t[r].Val = t[r].Val, t[l].Val
		}
	}
	return root
}

# Accepted solution for LeetCode #2415: Reverse Odd Levels of Binary Tree
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def reverseOddLevels(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        q = deque([root])
        i = 0
        while q:
            if i & 1:
                l, r = 0, len(q) - 1
                while l < r:
                    q[l].val, q[r].val = q[r].val, q[l].val
                    l, r = l + 1, r - 1
            for _ in range(len(q)):
                node = q.popleft()
                if node.left:
                    q.append(node.left)
                    q.append(node.right)
            i += 1
        return root

// Accepted solution for LeetCode #2415: Reverse Odd Levels of Binary Tree
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::collections::VecDeque;
use std::rc::Rc;
impl Solution {
    fn create_tree(vals: &Vec<Vec<i32>>, i: usize, j: usize) -> Option<Rc<RefCell<TreeNode>>> {
        if i == vals.len() {
            return None;
        }
        Some(Rc::new(RefCell::new(TreeNode {
            val: vals[i][j],
            left: Self::create_tree(vals, i + 1, j * 2),
            right: Self::create_tree(vals, i + 1, j * 2 + 1),
        })))
    }

    pub fn reverse_odd_levels(
        root: Option<Rc<RefCell<TreeNode>>>,
    ) -> Option<Rc<RefCell<TreeNode>>> {
        let mut queue = VecDeque::new();
        queue.push_back(root);
        let mut d = 0;
        let mut vals = Vec::new();
        while !queue.is_empty() {
            let mut val = Vec::new();
            for _ in 0..queue.len() {
                let mut node = queue.pop_front().unwrap();
                let mut node = node.as_mut().unwrap().borrow_mut();
                val.push(node.val);
                if node.left.is_some() {
                    queue.push_back(node.left.take());
                }
                if node.right.is_some() {
                    queue.push_back(node.right.take());
                }
            }
            if d % 2 == 1 {
                val.reverse();
            }
            vals.push(val);
            d += 1;
        }
        Self::create_tree(&vals, 0, 0)
    }
}

// Accepted solution for LeetCode #2415: Reverse Odd Levels of Binary Tree
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function reverseOddLevels(root: TreeNode | null): TreeNode | null {
    const q: TreeNode[] = [root];
    for (let i = 0; q.length > 0; ++i) {
        if (i % 2) {
            for (let l = 0, r = q.length - 1; l < r; ++l, --r) {
                [q[l].val, q[r].val] = [q[r].val, q[l].val];
            }
        }
        const nq: TreeNode[] = [];
        for (const { left, right } of q) {
            if (left) {
                nq.push(left);
                nq.push(right);
            }
        }
        q.splice(0, q.length, ...nq);
    }
    return root;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.