Most Frequent Subtree Sum

Description

Given the root of a tree, you are asked to find the most frequent subtree sum. 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). So what is the most frequent subtree sum value? If there is a tie, return all the values with the highest frequency in any order.

Examples 1 Input:

  5
 /  \
2   -3

return [2, -3, 4], since all the values happen only once, return all of them in any order.

Examples 2 Input:

  5
 /  \
2   -5

return [2], since 2 happens twice, however -5 only occur once.

Note: You may assume the sum of values in any subtree is in the range of 32-bit signed integer.

Solutions

Use DFS to find the frequencies of sums and then find the most frequent sum(s) in one pass.

class Solution {
public:
    vector<int> findFrequentTreeSum(TreeNode* root) {
        unordered_map<int, int> freqs;
        treeSum(root, freqs);
        vector<int> res;
        int max_freq = 0;
        for(auto it = freqs.begin(); it != freqs.end(); ++it) {
            if(it->second < max_freq) continue;
            if(it->second > max_freq){
                res.clear();
                max_freq = it->second;
            }
            res.push_back(it->first);
        }
        return res;
    }
    
    int treeSum(TreeNode *root, unordered_map<int, int> &freqs) {
        if(!root) return 0;
        int sum = treeSum(root->left, freqs) + treeSum(root->right, freqs) + root->val;
        freqs[sum]++;
        return sum;
    }
};

This method updates max_freq more number of times, but will not push temporary most frequent sums into res.

class Solution {
public:
    vector<int> findFrequentTreeSum(TreeNode* root) {
        unordered_map<int, int> freqs;
        int max_freq = 0;
        treeSum(root, freqs, max_freq);
        vector<int> res;
        for(auto it = freqs.begin(); it != freqs.end(); ++it)
            if(it->second == max_freq)
                res.push_back(it->first);
        return res;
    }
    
    int treeSum(TreeNode *root, unordered_map<int, int> &freqs, int &max_freq) {
        if(!root) return 0;
        int sum = treeSum(root->left, freqs, max_freq) + treeSum(root->right, freqs, max_freq) + root->val;
        max_freq = max(max_freq, ++freqs[sum]);
        return sum;
    }
};