Most Frequent Subtree Sum

Most Frequent Subtree Sum - 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).

Input & Output

Example 1 — Basic Tree

$ Input: root = [5,2,-3]

› Output: [2,-3,4]

💡 Note: Subtree sums: Node 2 has sum 2, Node -3 has sum -3, Node 5 has sum 5+2+(-3)=4. All appear once, so return all.

Example 2 — Duplicate Sums

$ Input: root = [5,2,-5]

› Output: [2]

💡 Note: Subtree sums: Node 2 has sum 2, Node -5 has sum -5, Node 5 has sum 5+2+(-5)=2. Sum 2 appears twice (most frequent), so return [2].

Example 3 — Single Node

$ Input: root = [1]

› Output: [1]

💡 Note: Only one subtree with sum 1, so it's the most frequent.

Constraints

The number of nodes in the tree is in the range [1, 10⁴]
-10⁵ ≤ Node.val ≤ 10⁵

Visualization

Tap to expand

Asked in

a Amazon 15 G Google 12 f Facebook 8

The key insight is to use post-order DFS traversal to calculate subtree sums bottom-up while tracking their frequencies. The optimal approach visits each node once, calculates its subtree sum from children, and maintains a frequency map. Time: O(n), Space: O(n).

Common Approaches

✓ Brute Force - Separate Sum and Count

⏱️ Time: O(n²) Space: O(n)

First traverse the tree to calculate sum for each possible subtree, store all sums in a list, then iterate through the list to count frequencies and find the maximum.

Optimized DFS with Frequency Tracking

⏱️ Time: O(n) Space: O(n)

Use post-order DFS to calculate subtree sums bottom-up in one pass, while simultaneously tracking frequency of each sum in a hash map.

Brute Force - Separate Sum and Count — Algorithm Steps

Step 1: For each node, calculate its subtree sum recursively
Step 2: Store all subtree sums in a list
Step 3: Count frequency of each sum
Step 4: Return sums with maximum frequency

Visualization

Tap to expand

Step-by-Step Walkthrough

Collect Nodes

Get all nodes in the tree

Calculate Sums

For each node, calculate its subtree sum

Count Frequencies

Count how often each sum appears

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MAX_NODES 1000

struct TreeNode {
    int val;
    struct TreeNode* left;
    struct TreeNode* right;
};

int sumsInOrder[MAX_NODES];
int sumsCount = 0;
int freqKeys[MAX_NODES];
int freqValues[MAX_NODES];
int freqCount = 0;

struct TreeNode* createNode(int val) {
    struct TreeNode* node = malloc(sizeof(struct TreeNode));
    node->val = val;
    node->left = NULL;
    node->right = NULL;
    return node;
}

void addToFreq(int sum) {
    for (int i = 0; i < freqCount; i++) {
        if (freqKeys[i] == sum) {
            freqValues[i]++;
            return;
        }
    }
    freqKeys[freqCount] = sum;
    freqValues[freqCount] = 1;
    freqCount++;
}

int getFreq(int sum) {
    for (int i = 0; i < freqCount; i++) {
        if (freqKeys[i] == sum) {
            return freqValues[i];
        }
    }
    return 0;
}

int dfs(struct TreeNode* node) {
    if (!node) {
        return 0;
    }
    
    int leftSum = dfs(node->left);
    int rightSum = dfs(node->right);
    int currentSum = node->val + leftSum + rightSum;
    
    sumsInOrder[sumsCount++] = currentSum;
    addToFreq(currentSum);
    
    return currentSum;
}

int* solution(struct TreeNode* root, int* returnSize) {
    *returnSize = 0;
    if (!root) return NULL;
    
    sumsCount = 0;
    freqCount = 0;
    
    dfs(root);
    
    // Find maximum frequency
    int maxFreq = 0;
    for (int i = 0; i < freqCount; i++) {
        if (freqValues[i] > maxFreq) {
            maxFreq = freqValues[i];
        }
    }
    
    // Return sums with max frequency in the order they were encountered
    int* result = malloc(MAX_NODES * sizeof(int));
    int seen[MAX_NODES];
    int seenCount = 0;
    
    for (int i = 0; i < sumsCount; i++) {
        int s = sumsInOrder[i];
        if (getFreq(s) == maxFreq) {
            int alreadySeen = 0;
            for (int j = 0; j < seenCount; j++) {
                if (seen[j] == s) {
                    alreadySeen = 1;
                    break;
                }
            }
            if (!alreadySeen) {
                result[(*returnSize)++] = s;
                seen[seenCount++] = s;
            }
        }
    }
    
    return result;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        if (strncmp(p, "null", 4) == 0) {
            arr[(*size)++] = -999999; // Use special value for null
            p += 4;
        } else {
            arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
        }
    }
}

struct TreeNode* buildTree(int* arr, int size) {
    if (size == 0 || arr[0] == -999999) return NULL;
    
    struct TreeNode* root = createNode(arr[0]);
    struct TreeNode* queue[MAX_NODES];
    int front = 0, rear = 0;
    queue[rear++] = root;
    int i = 1;
    
    while (front < rear && i < size) {
        struct TreeNode* node = queue[front++];
        
        if (i < size && arr[i] != -999999) {
            node->left = createNode(arr[i]);
            queue[rear++] = node->left;
        }
        i++;
        
        if (i < size && arr[i] != -999999) {
            node->right = createNode(arr[i]);
            queue[rear++] = node->right;
        }
        i++;
    }
    
    return root;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int arr[MAX_NODES];
    int size;
    parseArray(line, arr, &size);
    
    struct TreeNode* root = buildTree(arr, size);
    int returnSize;
    int* result = solution(root, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        if (i > 0) printf(",");
        printf("%d", result[i]);
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

O(n) for each of n nodes to calculate subtree sum

⚠ Quadratic Growth

Space Complexity

O(n)

List to store all n subtree sums plus recursion stack

⚡ Linearithmic Space

32.0K Views

Medium Frequency

~15 min Avg. Time

892 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Separate Sum and Count — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler