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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) {
                res.add(entry.getKey());
            }
        }
        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
    }
}

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