Most Frequent Subtree Sum solution and explanation
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The problem
Given the root of a binary tree, return the most frequent subtree sum. If there is a tie, return all the values with the highest frequency in any order.
The subtree sum of a node is defined as the sum of all the node values formed by the subtree rooted at that node (including the node itself).
Example 1:
Input: root = [5,2,-3] Output: [2,-3,4]
Example 2:
Input: root = [5,2,-5] Output: [2]
Constraints:
- The number of nodes in the tree is in the range
[1, 104]. -105 <= Node.val <= 105
Approach overview
Problem Overview: You receive the root of a binary tree and must compute the sum of every subtree. A subtree sum equals the value of the node plus the sums of its left and right subtrees. The task is to return the value(s) that appear most frequently among all subtree sums.
Approach 1: Post-order Traversal with Hash Map (O(n) time, O(n) space)
The most direct solution uses depth-first search with post-order traversal. Process the left and right children first, compute their subtree sums, then calculate the current node's sum as node.val + leftSum + rightSum. Store each computed sum in a hash table that tracks how many times each sum appears.
Post-order traversal is critical because a node’s subtree sum depends on its children. As DFS unwinds, update the frequency map and track the maximum occurrence count. After the traversal finishes, iterate through the map and collect all sums with the highest frequency. This approach touches each node exactly once, giving O(n) time complexity and O(n) space for the recursion stack and frequency map.
This method is efficient and straightforward. Most production and interview solutions follow this pattern because it combines computation and frequency tracking in a single traversal of the binary tree.
Approach 2: Pre-compute Subtree Sums then DFS Frequency Count (O(n) time, O(n) space)
This variant separates the computation into two logical phases. First perform a DFS to compute and store the subtree sum for every node. The recursion still follows post-order so that child sums are available before calculating the parent’s sum. Store each result in a list or array during traversal.
Once all subtree sums are collected, run a second pass that counts frequencies using a hash map. Track the maximum frequency while iterating over the stored sums. Finally, return all sums whose frequency matches that maximum.
The algorithm still runs in O(n) time because each node is processed a constant number of times. Space usage remains O(n) for the stored subtree sums and the frequency map. This structure can make debugging easier because the computation and counting phases are separated, but it requires slightly more memory and an additional pass.
Recommended for interviews: The post-order traversal with a hash map is the expected solution. It demonstrates a solid understanding of recursive tree processing and efficient frequency counting. The two-pass approach also works, but interviewers usually prefer the single DFS version because it computes subtree sums and tracks frequencies in one traversal.
Approach 1: Post-order Traversal with Hash Map
This approach involves using a post-order traversal to calculate the subtree sum for each node. We store each computed sum in a hash map or dictionary which keeps track of how many times each sum appears. After the traversal, we determine the maximum frequency of the sums and return the list of all sums that have this maximum frequency.
from collections import defaultdict
from typing import List
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: TreeNode) -> List[int]:
def postOrder(node):
if not node:
return 0
left_sum = postOrder(node.left)
right_sum = postOrder(node.right)
total_sum = node.val + left_sum + right_sum
sum_frequency[total_sum] += 1
return total_sum
sum_frequency = defaultdict(int)
postOrder(root)
max_freq = max(sum_frequency.values(), default=0)
return [s for s, freq in sum_frequency.items() if freq == max_freq]
# Test case
root = TreeNode(5, TreeNode(2), TreeNode(-3))
sol = Solution()
print(sol.findFrequentTreeSum(root)) # Output: [2, -3, 4]import java.util.*;
class TreeNode {
int val;
TreeNode left, right;
TreeNode(int x) {
val = x;
}
}
class Solution {
private Map<Integer, Integer> sumFrequency = new HashMap<>();
private int postOrder(TreeNode root) {
if (root == null) return 0;
int leftSum = postOrder(root.left);
int rightSum = postOrder(root.right);
int totalSum = root.val + leftSum + rightSum;
sumFrequency.put(totalSum, sumFrequency.getOrDefault(totalSum, 0) + 1);
return totalSum;
}
public int[] findFrequentTreeSum(TreeNode root) {
postOrder(root);
int maxFreq = Collections.max(sumFrequency.values());
List<Integer> result = new ArrayList<>();
for (Map.Entry<Integer, Integer> entry : sumFrequency.entrySet()) {
if (entry.getValue() == maxFreq) {
result.add(entry.getKey());
}
}
return result.stream().mapToInt(i -> i).toArray();
}
public static void main(String[] args) {
TreeNode root = new TreeNode(5);
root.left = new TreeNode(2);
root.right = new TreeNode(-3);
Solution sol = new Solution();
int[] result = sol.findFrequentTreeSum(root);
System.out.println(Arrays.toString(result));
}
}#include <iostream>
#include <unordered_map>
#include <vector>
using namespace std;
struct TreeNode {
int val;
TreeNode *left, *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
unordered_map<int, int> sumFrequency;
int postOrder(TreeNode* root) {
if (!root) return 0;
int leftSum = postOrder(root->left);
int rightSum = postOrder(root->right);
int totalSum = root->val + leftSum + rightSum;
sumFrequency[totalSum]++;
return totalSum;
}
vector<int> findFrequentTreeSum(TreeNode* root) {
postOrder(root);
int maxFreq = 0;
for (auto &entry : sumFrequency) {
if (entry.second > maxFreq) {
maxFreq = entry.second;
}
}
vector<int> result;
for (auto &entry : sumFrequency) {
if (entry.second == maxFreq) {
result.push_back(entry.first);
}
}
return result;
}
};
int main() {
TreeNode* root = new TreeNode(5);
root->left = new TreeNode(2);
root->right = new TreeNode(-3);
Solution sol;
vector<int> result = sol.findFrequentTreeSum(root);
for (int val : result) {
cout << val << " ";
}
return 0;
}class TreeNode {
constructor(val) {
this.val = val;
this.left = this.right = null;
}
}
class Solution {
constructor() {
this.sumFrequency = new Map();
}
postOrder(node) {
if (!node) return 0;
let leftSum = this.postOrder(node.left);
let rightSum = this.postOrder(node.right);
let totalSum = node.val + leftSum + rightSum;
this.sumFrequency.set(totalSum, (this.sumFrequency.get(totalSum) || 0) + 1);
return totalSum;
}
findFrequentTreeSum(root) {
this.postOrder(root);
let maxFreq = Math.max(...this.sumFrequency.values());
return [...this.sumFrequency.entries()].filter(([sum, freq]) => freq === maxFreq).map(([sum, freq]) => sum);
}
}
let root = new TreeNode(5);
root.left = new TreeNode(2);
root.right = new TreeNode(-3);
let solution = new Solution();
console.log(solution.findFrequentTreeSum(root)); // Output: [2, -3, 4]#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
typedef struct TreeNode {
int val;
struct TreeNode *left, *right;
} TreeNode;
typedef struct {
int *sums;
int size, capacity;
} ResultSet;
typedef struct {
int key, value;
} Pair;
typedef struct {
Pair *data;
int size;
} HashMap;
void initHashMap(HashMap *map) {
map->data = (Pair *)malloc(1000 * sizeof(Pair)); // dynamic max size
map->size = 0;
}
int findIndex(HashMap *map, int key) {
for (int i = 0; i < map->size; i++) {
if (map->data[i].key == key) {
return i;
}
}
return -1;
}
void put(HashMap *map, int key, int value) {
int index = findIndex(map, key);
if (index == -1) { // new key
map->data[map->size].key = key;
map->data[map->size].value = value;
map->size++;
} else { // existing key
map->data[index].value += value;
}
}
int get(HashMap *map, int key) {
int index = findIndex(map, key);
return index != -1 ? map->data[index].value : 0;
}
int postOrder(TreeNode *root, HashMap *map) {
if (!root) return 0;
int left = postOrder(root->left, map);
int right = postOrder(root->right, map);
int sum = root->val + left + right;
put(map, sum, 1);
return sum;
}
ResultSet findFrequentTreeSum(TreeNode* root) {
HashMap map;
initHashMap(&map);
postOrder(root, &map);
int maxFreq = 0;
for (int i = 0; i < map.size; i++) {
if (map.data[i].value > maxFreq) {
maxFreq = map.data[i].value;
}
}
ResultSet res;
res.size = 0;
res.capacity = 100;
res.sums = (int *)malloc(res.capacity * sizeof(int));
for (int i = 0; i < map.size; i++) {
if (map.data[i].value == maxFreq) {
if (res.size >= res.capacity) {
res.capacity *= 2;
res.sums = (int *)realloc(res.sums, res.capacity * sizeof(int));
}
res.sums[res.size++] = map.data[i].key;
}
}
return res;
}
int main() {
// Binary tree creation and test
TreeNode node1 = {5, NULL, NULL};
TreeNode node2 = {2, NULL, NULL};
TreeNode node3 = {-3, NULL, NULL};
node1.left = &node2;
node1.right = &node3;
ResultSet result = findFrequentTreeSum(&node1);
for (int i = 0; i < result.size; i++) {
printf("%d ", result.sums[i]);
}
free(result.sums);
return 0;
}using System;
using System.Collections.Generic;
using System.Linq;
public class TreeNode {
public int val;
public TreeNode left;
public TreeNode right;
public TreeNode(int x) { val = x; }
}
public class Solution {
private Dictionary<int, int> sumFrequency = new Dictionary<int, int>();
private int PostOrder(TreeNode node) {
if (node == null) return 0;
int leftSum = PostOrder(node.left);
int rightSum = PostOrder(node.right);
int totalSum = node.val + leftSum + rightSum;
if (sumFrequency.ContainsKey(totalSum)) {
sumFrequency[totalSum]++;
} else {
sumFrequency[totalSum] = 1;
}
return totalSum;
}
public int[] FindFrequentTreeSum(TreeNode root) {
PostOrder(root);
int maxFreq = sumFrequency.Values.Max();
return sumFrequency.Where(entry => entry.Value == maxFreq).Select(entry => entry.Key).ToArray();
}
}
public class Program {
public static void Main() {
TreeNode root = new TreeNode(5) {
left = new TreeNode(2),
right = new TreeNode(-3)
};
Solution solution = new Solution();
int[] result = solution.FindFrequentTreeSum(root);
Console.WriteLine(String.Join(", ", result));
}
}This Python solution uses a defaultdict to count the frequency of each subtree sum as they are computed in a post-order tree traversal. After collecting all sums, it finds the maximum frequency and gathers all sums with this frequency for the result.
Time Complexity: O(N), where N is the number of nodes because each node is visited once.
Space Complexity: O(N), for storing the subtree sums in a dictionary.
Approach 2: Pre-compute and DFS
This method involves pre-computing the subtree sums using a depth-first search (DFS). A map is used to record the number of occurrences of each sum, similar to the first method but differing in traversal style. Instead of post-order, a conventional DFS approach computes sums as the tree is traversed depth-first.
from collections import defaultdict
from typing import List
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: TreeNode) -> List[int]:
def dfs(node):
if not node:
return 0
left_sum = dfs(node.left)
right_sum = dfs(node.right)
total_sum = node.val + left_sum + right_sum
sum_frequency[total_sum] += 1
return total_sum
sum_frequency = defaultdict(int)
dfs(root)
max_freq = max(sum_frequency.values(), default=0)
return [s for s, freq in sum_frequency.items() if freq == max_freq]
# Test case
root = TreeNode(5, TreeNode(2), TreeNode(-3))
sol = Solution()
print(sol.findFrequentTreeSum(root))import java.util.*;
class TreeNode {
int val;
TreeNode left, right;
TreeNode(int x) { val = x; }
}
class Solution {
private Map<Integer, Integer> sumFrequency = new HashMap<>();
private int dfs(TreeNode root) {
if (root == null) return 0;
int leftSum = dfs(root.left);
int rightSum = dfs(root.right);
int totalSum = root.val + leftSum + rightSum;
sumFrequency.put(totalSum, sumFrequency.getOrDefault(totalSum, 0) + 1);
return totalSum;
}
public int[] findFrequentTreeSum(TreeNode root) {
dfs(root);
int maxFreq = Collections.max(sumFrequency.values());
List<Integer> result = new ArrayList<>();
for (Map.Entry<Integer, Integer> entry : sumFrequency.entrySet()) {
if (entry.getValue() == maxFreq) {
result.add(entry.getKey());
}
}
return result.stream().mapToInt(i -> i).toArray();
}
public static void main(String[] args) {
TreeNode root = new TreeNode(5);
root.left = new TreeNode(2);
root.right = new TreeNode(-3);
Solution sol = new Solution();
System.out.println(Arrays.toString(sol.findFrequentTreeSum(root)));
}
}#include <iostream>
#include <unordered_map>
#include <vector>
using namespace std;
struct TreeNode {
int val;
TreeNode *left, *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
unordered_map<int, int> freqMap;
int dfs(TreeNode* node) {
if (!node) return 0;
int leftSum = dfs(node->left);
int rightSum = dfs(node->right);
int sum = node->val + leftSum + rightSum;
freqMap[sum]++;
return sum;
}
public:
vector<int> findFrequentTreeSum(TreeNode* root) {
dfs(root);
vector<int> maxFreqSums;
int maxFreq = 0;
for (const auto& entry : freqMap) {
if (entry.second > maxFreq) {
maxFreq = entry.second;
maxFreqSums = {entry.first};
} else if (entry.second == maxFreq) {
maxFreqSums.push_back(entry.first);
}
}
return maxFreqSums;
}
};
int main() {
TreeNode* root = new TreeNode(5);
root->left = new TreeNode(2);
root->right = new TreeNode(-3);
Solution sol;
vector<int> result = sol.findFrequentTreeSum(root);
for (int val : result) {
cout << val << " ";
}
return 0;
}class TreeNode {
constructor(val) {
this.val = val;
this.left = this.right = null;
}
}
class Solution {
constructor() {
this.sumFrequency = new Map();
}
dfs(node) {
if (!node) return 0;
let leftSum = this.dfs(node.left);
let rightSum = this.dfs(node.right);
let totalSum = node.val + leftSum + rightSum;
this.sumFrequency.set(totalSum, (this.sumFrequency.get(totalSum) || 0) + 1);
return totalSum;
}
findFrequentTreeSum(root) {
this.dfs(root);
let maxFreq = Math.max(...this.sumFrequency.values());
return [...this.sumFrequency.entries()].filter(([sum, freq]) => freq === maxFreq).map(([sum, freq]) => sum);
}
}
let root = new TreeNode(5);
root.left = new TreeNode(2);
root.right = new TreeNode(-3);
let solution = new Solution();
console.log(solution.findFrequentTreeSum(root));#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_TREE_SIZE 10000
#define HASH_SIZE 20000
typedef struct TreeNode {
int val;
struct TreeNode *left, *right;
} TreeNode;
typedef struct ListNode {
int sum;
int count;
struct ListNode *next;
} ListNode;
ListNode* hashTable[HASH_SIZE];
int hash(int key) {
return abs(key) % HASH_SIZE;
}
void insertHash(int sum) {
int index = hash(sum);
ListNode *node = hashTable[index];
while (node != NULL) {
if (node->sum == sum) {
node->count++;
return;
}
node = node->next;
}
node = (ListNode*)malloc(sizeof(ListNode));
node->sum = sum;
node->count = 1;
node->next = hashTable[index];
hashTable[index] = node;
}
int dfs(TreeNode *root) {
if (!root) return 0;
int left_sum = dfs(root->left);
int right_sum = dfs(root->right);
int sum = root->val + left_sum + right_sum;
insertHash(sum);
return sum;
}
int* findFrequentTreeSum(TreeNode* root, int* returnSize) {
dfs(root);
int maxCount = 0;
int result[MAX_TREE_SIZE];
int idx = 0;
for (int i = 0; i < HASH_SIZE; i++) {
ListNode *node = hashTable[i];
while (node) {
if (node->count > maxCount) {
maxCount = node->count;
idx = 0;
result[idx++] = node->sum;
} else if (node->count == maxCount) {
result[idx++] = node->sum;
}
node = node->next;
}
}
int *output = (int*)malloc(idx * sizeof(int));
memcpy(output, result, idx * sizeof(int));
*returnSize = idx;
return output;
}
int main() {
TreeNode root = {5, NULL, NULL};
TreeNode left = {2, NULL, NULL};
TreeNode right = {-3, NULL, NULL};
root.left = &left;
root.right = &right;
int returnSize;
int *result = findFrequentTreeSum(&root, &returnSize);
for (int i = 0; i < returnSize; i++) {
printf("%d ", result[i]);
}
free(result);
return 0;
}using System;
using System.Collections.Generic;
using System.Linq;
public class TreeNode {
public int val;
public TreeNode left;
public TreeNode right;
public TreeNode(int x) { val = x; }
}
public class Solution {
private Dictionary<int, int> sumFrequency = new Dictionary<int, int>();
private int DFS(TreeNode node) {
if (node == null) return 0;
int leftSum = DFS(node.left);
int rightSum = DFS(node.right);
int totalSum = node.val + leftSum + rightSum;
if (sumFrequency.ContainsKey(totalSum)) {
sumFrequency[totalSum]++;
} else {
sumFrequency[totalSum] = 1;
}
return totalSum;
}
public int[] FindFrequentTreeSum(TreeNode root) {
DFS(root);
int maxFreq = sumFrequency.Values.Max();
return sumFrequency.Where(entry => entry.Value == maxFreq).Select(entry => entry.Key).ToArray();
}
public static void Main() {
TreeNode root = new TreeNode(5) {
left = new TreeNode(2),
right = new TreeNode(-3)
};
Solution solution = new Solution();
int[] result = solution.FindFrequentTreeSum(root);
Console.WriteLine(String.Join(", ", result));
}
}Implementing DFS, this Python approach iteratively computes subtree totals from each node, recording their frequencies in a dictionary, and determines which sums occur most often for its resulting output.
Time Complexity: O(N), as each node is explored.
Space Complexity: O(N), preserved in hashmap.
Approach 3: Hash Table + DFS
We can use a hash table cnt to record the frequency of each subtree sum. Then, we use depth-first search (DFS) to traverse the entire tree, calculate the sum of elements for each subtree, and update cnt.
Finally, we traverse cnt to find all subtree sums that appear most frequently.
The time complexity is O(n), and the space complexity is O(n). Here, n is the number of nodes in the 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 findFrequentTreeSum(root: TreeNode | null): number[] {
const cnt = new Map<number, number>();
let mx = 0;
const dfs = (root: TreeNode | null): number => {
if (!root) {
return 0;
}
const { val, left, right } = root;
const s = val + dfs(left) + dfs(right);
cnt.set(s, (cnt.get(s) ?? 0) + 1);
mx = Math.max(mx, cnt.get(s)!);
return s;
};
dfs(root);
return Array.from(cnt.entries())
.filter(([_, c]) => c === mx)
.map(([s, _]) => s);
}/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func findFrequentTreeSum(root *TreeNode) (ans []int) {
cnt := map[int]int{}
var mx int
var dfs func(*TreeNode) int
dfs = func(root *TreeNode) int {
if root == nil {
return 0
}
s := root.Val + dfs(root.Left) + dfs(root.Right)
cnt[s]++
mx = max(mx, cnt[s])
return s
}
dfs(root)
for k, v := range cnt {
if v == mx {
ans = append(ans, k)
}
}
return
}// 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::HashMap;
use std::rc::Rc;
impl Solution {
pub fn find_frequent_tree_sum(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
fn dfs(root: Option<Rc<RefCell<TreeNode>>>, cnt: &mut HashMap<i32, i32>) -> i32 {
if let Some(node) = root {
let l = dfs(node.borrow().left.clone(), cnt);
let r = dfs(node.borrow().right.clone(), cnt);
let s = l + r + node.borrow().val;
*cnt.entry(s).or_insert(0) += 1;
s
} else {
0
}
}
let mut cnt = HashMap::new();
dfs(root, &mut cnt);
let mx = cnt.values().cloned().max().unwrap_or(0);
cnt.into_iter()
.filter(|&(_, v)| v == mx)
.map(|(k, _)| k)
.collect()
}
}Complexity comparison
| Approach | Complexity |
|---|---|
| Post-order Traversal with Hash Map | Time Complexity: O(N), where N is the number of nodes because each node is visited once. |
| Pre-compute and DFS | Time Complexity: O(N), as each node is explored. |
| Hash Table + DFS | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Post-order Traversal with Hash Map | O(n) | O(n) | General case. Best interview solution because subtree sums and frequency tracking happen in one DFS. |
| Pre-compute Subtree Sums + DFS Counting | O(n) | O(n) | Useful when separating computation and analysis phases for debugging or clarity. |
Frequently asked questions
Is Most Frequent Subtree Sum easy or hard?
Most Frequent Subtree Sum is rated Medium difficulty. The recursion itself is straightforward, but the challenge comes from correctly computing subtree sums using post-order traversal and tracking frequencies efficiently.
Most Frequent Subtree Sum Python/Java solution
Python and Java implementations typically use recursive DFS with a hash map or dictionary to track sum frequencies. Python uses a dictionary or collections.Counter, while Java uses HashMap<Integer, Integer>. Both achieve O(n) time and O(n) space complexity.
How to solve Most Frequent Subtree Sum in O(n)?
Perform a post-order depth-first traversal. For each node, compute the subtree sum using the sums of its left and right children. Store the frequency of each sum in a hash map and track the maximum frequency while traversing. After DFS completes, return all sums that match the highest frequency.
What is the best approach for Most Frequent Subtree Sum?
The best approach uses post-order DFS with a hash map to track frequencies of subtree sums. Each node computes its subtree sum after visiting its children, and the sum is recorded in the map. This processes the tree in a single traversal with O(n) time and O(n) space complexity.
Is Most Frequent Subtree Sum asked at Google/Amazon/Meta?
Binary tree DFS problems with subtree computations frequently appear in interviews at companies like Amazon, Meta, and Google. Variants involving subtree sums, subtree properties, or frequency counting are common because they test recursion, tree traversal, and hash map usage.
What data structure is used in Most Frequent Subtree Sum?
The solution relies on a hash table (hash map) to store the frequency of each subtree sum. The algorithm also uses recursion or an explicit stack for depth-first traversal of the binary tree.
What is the time complexity of Most Frequent Subtree Sum?
The optimal solution runs in O(n) time where n is the number of nodes in the binary tree. Each node is visited exactly once during the DFS traversal, and hash map updates occur in constant time on average.