algoadvance

Given the root of a binary tree, return the most frequent subtree sum. A subtree sum is defined as the sum of all the node values in that subtree, including the root. If there is a tie in the frequency of the sums, you should return all of them.

Clarifying Questions:

1. What is the structure of the tree nodes? The tree nodes are likely structured using a standard TreeNode class with attributes for value, left child, and right child.

2. Can the tree contain negative values? Yes, the tree can contain both positive and negative values.

3. Is there a limit on the tree’s size? No explicit limit is given, but it should be reasonable to handle in memory.

4. Would the tree always contain at least one node? Yes, per the problem definition, there is always at least the root node.

Strategy:

1. Use Depth-First Search (DFS) to traverse the tree.
2. At each node, compute the sum of its subtree (including itself).
3. Use a dictionary to keep track of the frequency of each subtree sum.
4. After the DFS traversal, determine the highest frequency of encountered subtree sums.
5. Collect all subtree sums that occur with the highest frequency and return them.

Code:

from collections import defaultdict
from typing import List, Optional

# 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 findFrequentTreeSum(self, root: Optional[TreeNode]) -> List[int]:
        if not root:
            return []
        
        sum_count = defaultdict(int)
        
        def dfs(node):
            if not node:
                return 0
            # Calculate the subtree sum
            left_sum = dfs(node.left)
            right_sum = dfs(node.right)
            subtree_sum = node.val + left_sum + right_sum
            # Record the sum in the frequency map
            sum_count[subtree_sum] += 1
            return subtree_sum

        # Initiate the DFS traversal from the root
        dfs(root)
        
        # Find the maximum frequency of the subtree sums
        max_frequency = max(sum_count.values())
        
        # Collect all the sums with the maximum frequency
        most_frequent_sums = [s for s, freq in sum_count.items() if freq == max_frequency]
        
        return most_frequent_sums

Time Complexity:

• Time Complexity: O(N) where N is the number of nodes in the tree. This is because we visit each node exactly once during the DFS traversal.
• Space Complexity: O(N) because in the worst case, we could be storing N/2 subtree sums for a complete binary tree. Additionally, the recursion stack could have up to N calls in the worst case of a skewed tree.

This solution is efficient and should work well within typical constraints for binary tree problems.

Try our interview co-pilot at AlgoAdvance.com

• Got blindsided by a question you didn’t expect?

• Spend too much time studying?

• Or simply don’t have the time to go over all 3000 questions?