508. Most Frequent Subtree Sum

最新推荐文章于 2025-07-29 14:24:32 发布

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最新推荐文章于 2025-07-29 14:24:32 发布

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分类专栏： LeetCode Tree Hash Table 文章标签： LeetCode Tree Hash Table Most Frequent Subtre

本文链接：https://blog.youkuaiyun.com/zshouyi/article/details/61417206

LeetCode 同时被 3 个专栏收录

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本文介绍了一种算法，用于找到给定树中出现频率最高的子树和，并提供了详细的实现步骤及示例。通过使用哈希映射记录每个子树和的出现次数，再运用桶排序找出最高频次的子树和。

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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.

建立hashmap，存下来所有子树的subsum和subsum出现的次数count。之后用bucket sort找出出现次数最多的subsum。代码如下：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
    int count = 0;
    
    public int[] findFrequentTreeSum(TreeNode root) {
        int[] res = new int[0];
        if (root == null) {
            return res;
        }
        helper(root);
        List<Integer>[] list = new List[count + 1];
        for (int num: map.keySet()) {
            int sumcount = map.get(num);
            if (list[sumcount] == null) {
                list[sumcount] = new ArrayList<Integer>();
            }
            list[sumcount].add(num);
        }
        for (int i = count; i >= 0; i --) {
            if (list[i] != null) {
                res = new int[list[i].size()];
                for (int j = 0; j < list[i].size(); j ++) {
                    res[j] = list[i].get(j);
                }
                break;
            }
        }
        return res;
    }
    
    private int helper(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int subsum = helper(root.left) + helper(root.right) + root.val;
        map.put(subsum, map.getOrDefault(subsum, 0) + 1);
        count ++;
        return subsum;
    }
}