How do you solve Average of Levels in Binary Tree on LeetCode?

Average of Levels in Binary Tree (LeetCode #637) 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 methods is crucial.

What is the time complexity of Average of Levels in Binary Tree?

The optimal solution for Average of Levels in Binary Tree runs in O(n) time and uses O(n) space. This complexity is due to visiting each node once, and the space complexity arises from the queue used for BFS.

What is the brute force solution for Average of Levels in Binary Tree?

The brute force approach for Average of Levels in Binary Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach calculates the average of each level by traversing the entire tree for each level. This is straightforward but inefficient as it repeatedly visits nodes. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Average of Levels in Binary Tree?

Average of Levels in Binary Tree uses the following concepts: Tree, Depth-First Search, Breadth-First Search, Binary Tree. The recommended patterns to study are: Breadth-First Search, Level Order Traversal.

What are the interview tips for Average of Levels in Binary Tree?

Clarify the problem statement before coding. Discuss your thought process and approach with the interviewer. Consider edge cases, like a single-node tree.

What are common mistakes when solving Average of Levels in Binary Tree?

Not using a queue for BFS, leading to incorrect level averages. Forgetting to handle null children in the tree.

#637

Average of Levels in Binary Tree

Easy

TreeDepth-First SearchBreadth-First SearchBinary TreeBreadth-First SearchLevel Order Traversal

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 approach uses a breadth-first search (BFS) to traverse the tree level by level, calculating the average in a single pass for each level.

⚙️

Algorithm

4 steps

1Step 1: Initialize a queue and add the root node.
2Step 2: While the queue is not empty, process all nodes at the current level.
3Step 3: For each node, add its value to a sum and enqueue its children.
4Step 4: After processing all nodes at the current level, calculate the average and store it.

solution.py25 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
8class Solution:
9    def averageOfLevels(self, root: TreeNode):
10        if not root:
11            return []
12        result = []
13        queue = [root]
14        while queue:
15            level_sum = 0
16            count = len(queue)
17            for _ in range(count):
18                node = queue.pop(0)
19                level_sum += node.val
20                if node.left:
21                    queue.append(node.left)
22                if node.right:
23                    queue.append(node.right)
24            result.append(level_sum / count)
25        return result

ℹ

Complexity note: This complexity is due to visiting each node once, and the space complexity arises from the queue used for BFS.

1Understanding tree traversal methods is crucial.
2BFS is often more efficient for level-based problems.

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