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.

Tips:

其实是一个后序遍历，先计算左右两边的subtree sum，用map，全局变量maxFreq和count记录每次的sum，这个sum出现的频率，和最高频率以及最高频率出现的次数。最后遍历map即可得到答案。

Code：

public class Solution {
    int maxFreq = 0;
    int count = 0;
    public int[] findFrequentTreeSum(TreeNode root) {
        if (root == null) return new int[0];
        Map<Integer, Integer> map = new HashMap<>();
        helper(root, map);
        int[] res = new int[count];
        int k = 0;
        for (int i : map.keySet()) {
            if (map.get(i) == maxFreq) res[k++] = i;
        }
        return res;
    }

    private int helper(TreeNode root, Map<Integer, Integer> map) {
        if (root == null) return 0;
        int left = helper(root.left, map);
        int right = helper(root.right, map);
        int sum = root.val + left + right;
        map.put(sum, map.getOrDefault(sum, 0) + 1);
        if (map.get(sum) > maxFreq) {
            maxFreq = map.get(sum);
            count = 1;
        } else if (map.get(sum) == maxFreq) count++;
        return sum;
    }
}