508. Most Frequent Subtree Sum

Description

Given the root of a binary tree, return the most frequent subtree sum. If there is a tie, return all the values with the highest frequency in any order.

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).

Example 1:

Input: root = [5,2,-3]
Output: [2,-3,4]

Example 2:

Input: root = [5,2,-5]
Output: [2]

Constraints:

The number of nodes in the tree is in the range [1, 10⁴].
-10⁵ <= Node.val <= 10⁵

Solutions

Solution 1: Hash Table + DFS

We can use a hash table $\textit{cnt}$ to record the frequency of each subtree sum. Then, we use depth-first search (DFS) to traverse the entire tree, calculate the sum of elements for each subtree, and update $\textit{cnt}$.

Finally, we traverse $\textit{cnt}$ to find all subtree sums that appear most frequently.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary tree.

Python3JavaC++GoTypeScriptRust

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def findFrequentTreeSum(self, root: Optional[TreeNode]) -> List[int]:
        def dfs(root: Optional[TreeNode]) -> int:
            if root is None:
                return 0
            l, r = dfs(root.left), dfs(root.right)
            s = l + r + root.val
            cnt[s] += 1
            return s

        cnt = Counter()
        dfs(root)
        mx = max(cnt.values())
        return [k for k, v in cnt.items() if v == mx]

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private Map<Integer, Integer> cnt = new HashMap<>();
    private int mx;

    public int[] findFrequentTreeSum(TreeNode root) {
        dfs(root);
        List<Integer> ans = new ArrayList<>();
        for (var e : cnt.entrySet()) {
            if (e.getValue() == mx) {
                ans.add(e.getKey());
            }
        }
        return ans.stream().mapToInt(i -> i).toArray();
    }

    private int dfs(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int s = root.val + dfs(root.left) + dfs(root.right);
        mx = Math.max(mx, cnt.merge(s, 1, Integer::sum));
        return s;
    }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> findFrequentTreeSum(TreeNode* root) {
        unordered_map<int, int> cnt;
        int mx = 0;
        function<int(TreeNode*)> dfs = [&](TreeNode* root) -> int {
            if (!root) {
                return 0;
            }
            int s = root->val + dfs(root->left) + dfs(root->right);
            mx = max(mx, ++cnt[s]);
            return s;
        };
        dfs(root);
        vector<int> ans;
        for (const auto& [k, v] : cnt) {
            if (v == mx) {
                ans.push_back(k);
            }
        }
        return ans;
    }
};