How do you solve Reverse Odd Levels of Binary Tree on LeetCode?

Reverse Odd Levels of Binary Tree (LeetCode #2415) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding tree traversal is crucial.

What is the brute force solution for Reverse Odd Levels of Binary Tree?

The brute force approach for Reverse Odd Levels of Binary Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves traversing the tree level by level, collecting values from the odd levels, and then reversing those values before placing them back into the tree. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Reverse Odd Levels of Binary Tree?

Reverse Odd Levels of Binary Tree uses the following concepts: Tree, Depth-First Search, Breadth-First Search, Binary Tree. The recommended patterns to study are: Depth-First Search, Breadth-First Search.

What are the interview tips for Reverse Odd Levels of Binary Tree?

Explain your thought process clearly. Consider edge cases, like trees with only one level. Practice tree traversal techniques before the interview.

What are common mistakes when solving Reverse Odd Levels of Binary Tree?

Not keeping track of the current level correctly. Forgetting to reverse values after collecting them.

#2415

Reverse Odd Levels of Binary Tree

Medium

TreeDepth-First SearchBreadth-First SearchBinary TreeDepth-First SearchBreadth-First Search

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a depth-first search (DFS) to traverse the tree and directly reverse the values at odd levels as we go. This avoids the need for additional space to store values and allows us to reverse in a single pass.

⚙️

Algorithm

3 steps

1Step 1: Use DFS to traverse the tree, keeping track of the current level.
2Step 2: At odd levels, collect the values in a list.
3Step 3: Once the entire level is collected, reverse the list and assign the values back to the nodes.

solution.py26 lines

1# Full working Python code
2class TreeNode:
3    def __init__(self, val=0, left=None, right=None):
4        self.val = val
5        self.left = left
6        self.right = right
7
8def reverseOddLevels(root):
9    def dfs(node, level):
10        if not node:
11            return
12        if level % 2 == 1:
13            if level not in odd_levels:
14                odd_levels[level] = []
15            odd_levels[level].append(node)
16        dfs(node.left, level + 1)
17        dfs(node.right, level + 1)
18
19    odd_levels = {}
20    dfs(root, 0)
21    for nodes in odd_levels.values():
22        values = [node.val for node in nodes]
23        values.reverse()
24        for i in range(len(nodes)):
25            nodes[i].val = values[i]
26    return root

ℹ

Complexity note: The time complexity is O(n) because we visit each node exactly once. The space complexity is O(n) due to the storage of nodes at odd levels in a map.

1Understanding tree traversal is crucial.
2Reversing values in place can save space.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.