[leetcode] 508. Most Frequent Subtree Sum

最新推荐文章于 2022-11-24 17:07:40 发布

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本文介绍了一种通过递归计算二叉树中最常出现的子树和的方法，并使用哈希表记录每个子树和出现的频率，最后找出出现频率最高的子树和值。

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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.

这道题是统计出现频率最高的Subtree Sum，题目难度为Medium。

可以通过递归计算Subtree Sum，相信大家都明白，就不详细说明了。这里用Hash Table来存储Suntree Sum出现的次数，在计算出Subtree Sum之后，更新Hash Table中次数。遍历Hash Table找出最大次数，然后就可以得出最终结果了。具体代码：

class Solution {
    int getSubtreeSum(unordered_map<int, int>& cnt, TreeNode* node) {
        if(node == NULL) return 0;
        int sum = node->val;
        sum += getSubtreeSum(cnt, node->left);
        sum += getSubtreeSum(cnt, node->right);
        ++cnt[sum];
        return sum;
    }
public:
    vector<int> findFrequentTreeSum(TreeNode* root) {
        unordered_map<int, int> cnt;
        vector<int> ret;
        int maxFreq = INT_MIN;
        
        getSubtreeSum(cnt, root);
        
        for(auto p:cnt) {
            if(p.second > maxFreq) {
                maxFreq = p.second;
            }
        }
        for(auto p:cnt) {
            if(p.second == maxFreq) {
                ret.push_back(p.first);
            }
        }
        
        return ret;
    }
};