508 Most Frequent Subtree Sum

508. Most Frequent Subtree Sum

1. Question

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.

2. Implementation

思路：这题主要是通过post order travsersal求得subtree的 sum，然后通过hashmap存储每个subtree sum的frequency

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int[] findFrequentTreeSum(TreeNode root) {
        if (root == null) {
            return new int[0];
        }

        int[] max = new int[1];
        Map<Integer, Integer> map = new HashMap();
        findSubTreeSumByDFS(root, map, max);

        List<Integer> list = new ArrayList();

        for (int key : map.keySet()) {
            if (map.get(key) == max[0]) {
                list.add(key);
            }
        }

        int[] res = new int[list.size()];

        for (int i = 0; i < list.size(); i++) {
            res[i] = list.get(i);
        }
        return res;
    }

    public int findSubTreeSumByDFS(TreeNode node, Map<Integer, Integer> map, int[] max) {
        if (node == null) {
            return 0;
        }

        int leftSum = findSubTreeSumByDFS(node.left, map, max);
        int rightSum = findSubTreeSumByDFS(node.right, map, max);
        int sum = node.val + leftSum + rightSum;
        map.put(sum, map.getOrDefault(sum, 0) + 1);
        max[0] = Math.max(max[0], map.get(sum));

        return sum;
    }
}

3. Time & Space Complexity

时间复杂度为O(n), n为树里所有节点的个数，空间复杂度为O(m), m为所有可能的subtree sum