Most Frequent Subtree Sum问题及解法

最新推荐文章于 2018-06-05 23:37:10 发布

原创最新推荐文章于 2018-06-05 23:37:10 发布 · 264 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#LeetCode Medium

LeetCode算法练习（Medium）专栏收录该内容

238 篇文章

订阅专栏

本文介绍了一种算法，用于解决寻找二叉树中最频繁出现的子树和的问题。通过递归地计算每个节点的子树和，并使用哈希表记录各子树和的出现频率，最终找出出现频率最高的子树和。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

问题描述：

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.

示例：
Input:

  5
 /  \
2   -3

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

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.

问题分析：

将每棵树及子树的和都保存到hash表中，并记录出现次数最多的和，最后遍历hash表，即可找到答案。

过程详见代码：

/**
 * 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:
    int maxi;
	vector<int> findFrequentTreeSum(TreeNode* root) {
		unordered_map<int, int> m;
		vector<int> res;
        maxi = 0;
		bl(m, root);
		for (auto c : m)
		{
			if (c.second == maxi)
				res.emplace_back(c.first);
		}
		return res;
	}

	int bl(unordered_map<int, int>& m, TreeNode* root)
	{
		if (root == nullptr) return 0;
		int sum = bl(m, root->left) + bl(m, root->right) + root->val;
		if (!m.count(sum)) m[sum] = 1;
		else m[sum]++;
		maxi = max(maxi, m[sum]);
		return sum;
	}
    
};