508. Most Frequent Subtree Sum

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508. Most Frequent Subtree Sum

Description

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:

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

Example 2:

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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

Hints/Notes

  • N/A

Solution

Language: C++

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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:
    unordered_map<int, int> m;

    vector<int> findFrequentTreeSum(TreeNode* root) {
        traverse(root);
        int maxCount = -1;
        for (auto it : m) {
            maxCount = max(maxCount, it.second);
        }
        vector<int> res;
        for (auto it : m) {
            if (it.second == maxCount) {
                res.push_back(it.first);
            }
        }
        return res;
    }

    int traverse(TreeNode* root) {
        if (!root) {
            return 0;
        }
        int val = root->val;

        val += traverse(root->left);
        val += traverse(root->right);

        m[val]++;
        return val;
    }
};