[LeetCode] Most Frequent Subtree Sum

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.

找出出现次数最多的子树和。

1、使用map来存储子树和及其出现的次数

2、使用dfs来处理每一个节点对应的子树和。

3、遍历map来找出出现次数最多的子树和。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> findFrequentTreeSum(TreeNode* root) {
        vector<int> res;
        unordered_map<int, int> m;
        dfs(root, m);
        int maxCnt = 0;
        for (auto& x : m) {
            if (x.second > maxCnt) {
                res.clear();
                res.push_back(x.first);
                maxCnt = x.second;
            }
            else if (x.second == maxCnt) {
                res.push_back(x.first);
            }
        }
        return res;
    }
    
    int dfs(TreeNode* node, unordered_map<int, int>& m) {
        if (node == nullptr) {
            return 0;
        }
        int l = dfs(node->left, m);
        int r = dfs(node->right, m);
        int sum = node->val + l + r;
        m[sum]++;
        return sum;
    }
};
// 19 ms