Leetcode 508. Most Frequent Subtree Sum

题目链接

https://leetcode.com/problems/most-frequent-subtree-sum/description/

题目描述

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.

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.

题解

递归遍历每一个节点，计算子树的所有节点的和，然后用一个map存起来，如果当前sum出现的次数最大，那就清空已有的sum；如果当前sum的次数和已有的sum个数相等，就把该sum添加进数组，并保存在map里面。

代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private int maxCount = 0;
    public int[] findFrequentTreeSum(TreeNode root) {
        Map<Integer, Integer> map = new HashMap<>();
        List<Integer> list = new ArrayList<>();
        push(map, root, list);
        int[] a = new int[list.size()];
        for (int i = 0; i < list.size(); i++) {
            a[i] = list.get(i);
        }
        return a;
    }
    public int push(Map<Integer, Integer> map, TreeNode root, List<Integer> list) {
        if (root == null) {
            return 0;
        }
        int left = push(map, root.left, list);
        int right = push(map, root.right, list);
        int sum = left + right + root.val;
        int count = map.getOrDefault(sum, 0) + 1;
        if (count > maxCount) {
            list.clear();
            list.add(sum);
            maxCount = count;
        } else if (count == maxCount) {
            list.add(sum);
        }
        map.put(sum, count);
        return sum;
    }
}