How do you solve Most Frequent Subtree Sum on LeetCode?

Most Frequent Subtree Sum (LeetCode #508) 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: Subtree sums can be efficiently calculated using depth-first search.

What is the brute force solution for Most Frequent Subtree Sum?

The brute force approach for Most Frequent Subtree Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach calculates the subtree sum for each node and keeps track of the frequency of each sum. This is straightforward but inefficient, as it recalculates sums multiple times. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Most Frequent Subtree Sum?

Most Frequent Subtree Sum uses the following concepts: Hash Table, Tree, Depth-First Search, Binary Tree. The recommended patterns to study are: Hash Map, Depth-First Search.

What are the interview tips for Most Frequent Subtree Sum?

Explain your thought process clearly while coding. Discuss the time and space complexity of your approach. Consider edge cases and how your solution handles them.

What are common mistakes when solving Most Frequent Subtree Sum?

Not using a single traversal for subtree sums, leading to O(n²) complexity. Failing to handle edge cases like empty trees.

#508

Most Frequent Subtree Sum

Medium

Hash TableTreeDepth-First SearchBinary TreeHash MapDepth-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 single traversal of the tree to calculate subtree sums and their frequencies. This is efficient because it avoids redundant calculations by leveraging depth-first search.

⚙️

Algorithm

3 steps

1Step 1: Initialize a dictionary to count subtree sums and a variable to track the maximum frequency.
2Step 2: Perform a depth-first search to calculate subtree sums for each node while updating the frequency count.
3Step 3: After the traversal, identify the sums with the maximum frequency and return them.

solution.py27 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
8from collections import defaultdict
9
10def findFrequentTreeSum(root):
11    if not root:
12        return []
13
14    sum_count = defaultdict(int)
15    max_freq = 0
16
17    def subtree_sum(node):
18        nonlocal max_freq
19        if not node:
20            return 0
21        total = node.val + subtree_sum(node.left) + subtree_sum(node.right)
22        sum_count[total] += 1
23        max_freq = max(max_freq, sum_count[total])
24        return total
25
26    subtree_sum(root)
27    return [s for s, freq in sum_count.items() if freq == max_freq]

ℹ

Complexity note: The time complexity is O(n) because we traverse each node exactly once. The space complexity is O(n) due to the storage of subtree sums in the dictionary.

1Subtree sums can be efficiently calculated using depth-first search.
2Using a hash map allows for quick frequency counting.

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