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

• /**
* 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> counter;
private int mx;

public int[] findFrequentTreeSum(TreeNode root) {
counter = new HashMap<>();
mx = Integer.MIN_VALUE;
dfs(root);
List<Integer> res = new ArrayList<>();
for (Map.Entry<Integer, Integer> entry : counter.entrySet()) {
if (entry.getValue() == mx) {
}
}
int[] ans = new int[res.size()];
for (int i = 0; i < res.size(); ++i) {
ans[i] = res.get(i);
}
return ans;
}

private int dfs(TreeNode root) {
if (root == null) {
return 0;
}
int s = root.val + dfs(root.left) + dfs(root.right);
counter.put(s, counter.getOrDefault(s, 0) + 1);
mx = Math.max(mx, counter.get(s));
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:
unordered_map<int, int> counter;
int mx = 0;

vector<int> findFrequentTreeSum(TreeNode* root) {
mx = INT_MIN;
dfs(root);
vector<int> ans;
for (auto& entry : counter)
if (entry.second == mx)
ans.push_back(entry.first);
return ans;
}

int dfs(TreeNode* root) {
if (!root) return 0;
int s = root->val + dfs(root->left) + dfs(root->right);
++counter[s];
mx = max(mx, counter[s]);
return s;
}
};

• # 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: TreeNode) -> List[int]:
def dfs(root):
if root is None:
return 0
left, right = dfs(root.left), dfs(root.right)
s = root.val + left + right
counter[s] += 1
return s

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


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func findFrequentTreeSum(root *TreeNode) []int {
counter := make(map[int]int)
mx := 0
var dfs func(root *TreeNode) int
dfs = func(root *TreeNode) int {
if root == nil {
return 0
}
s := root.Val + dfs(root.Left) + dfs(root.Right)
counter[s]++
if mx < counter[s] {
mx = counter[s]
}
return s
}
dfs(root)
var ans []int
for k, v := range counter {
if v == mx {
ans = append(ans, k)
}
}
return ans
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

function findFrequentTreeSum(root: TreeNode | null): number[] {
const map = new Map<number, number>();
let max = 0;
const dfs = (root: TreeNode | null) => {
if (root == null) {
return 0;
}
const { val, left, right } = root;
const sum = val + dfs(left) + dfs(right);
map.set(sum, (map.get(sum) ?? 0) + 1);
max = Math.max(max, map.get(sum));
return sum;
};
dfs(root);
const res = [];
for (const [k, v] of map) {
if (v === max) {
res.push(k);
}
}
return res;
}


• // Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
use std::collections::HashMap;
impl Solution {
fn dfs(
root: &Option<Rc<RefCell<TreeNode>>>,
map: &mut HashMap<i32, i32>,
max: &mut i32
) -> i32 {
if root.is_none() {
return 0;
}
let node = root.as_ref().unwrap().borrow();
let sum = node.val + Self::dfs(&node.left, map, max) + Self::dfs(&node.right, map, max);
map.insert(sum, map.get(&sum).unwrap_or(&0) + 1);
*max = (*max).max(map[&sum]);
sum
}

pub fn find_frequent_tree_sum(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
let mut map = HashMap::new();
let mut max = 0;
let mut res = Vec::new();
Self::dfs(&root, &mut map, &mut max);
for (k, v) in map.into_iter() {
if v == max {
res.push(k);
}
}
res
}
}