Most Frequent Subtree Sum Problem

Description

LeetCode Problem 508.

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:

``````1
2
Input: root = [5,2,-3]
Output: [2,-3,4]
``````

Example 2:

``````1
2
Input: root = [5,2,-5]
Output: [2]
``````

Constraints:

• The number of nodes in the tree is in the range [1, 10^4].
• -10^5 <= Node.val <= 10^5

Sample C++ Code

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/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<int> findFrequentTreeSum(TreeNode* root) {
unordered_map<int,int> counts;
int maxCount = 0;
countSubtreeSums(root, counts, maxCount);

vector<int> maxSums;
for(const auto& x :  counts){
if(x.second == maxCount) maxSums.push_back(x.first);
}
return maxSums;
}

int countSubtreeSums(TreeNode *r, unordered_map<int,int> &counts, int& maxCount){
if(r == nullptr) return 0;
int sum = r->val;
sum += countSubtreeSums(r->left, counts, maxCount);
sum += countSubtreeSums(r->right, counts, maxCount);
++counts[sum];
maxCount = max(maxCount, counts[sum]);
return sum;
}
};
``````