508. 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.

Solution:

The idea is to perform a bottom-up traversal of the tree and calculate the subtree sum of each nodes.

The frequency of the sum is stored in a HashMap.

After this pre-computation, we go through the HashMap and find the maximum frequency.

Thus, we go through the HashMap one more time and put all the keys that map to the maximum frequency to the result list.



Code:

/**
 * 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<>();
    public int[] findFrequentTreeSum(TreeNode root) {
        if (root == null) {
            return new int[0];
        }
        helper(root);
        int max = Integer.MIN_VALUE;
        for (int val : map.values()) {
            max = Math.max(max, val);
        }
        List<Integer> list = new ArrayList<>();
        for (int key : map.keySet()) {
            if (map.get(key) == max) {
                list.add(key);
            }
        }
        int[] res = new int[list.size()];
        int i = 0;
        for (int val : list) {
            res[i++] = val;
        }
        return res;
    }
    public int helper(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int left = helper(root.left);
        int right = helper(root.right);
        int sum = left + right + root.val;
        if (!map.containsKey(sum)) {
            map.put(sum, 1);
        }
        else {
            map.put(sum, map.get(sum) + 1);
        }
        return sum;
    }
}