Question

508. 出现次数最多的子树元素和

English Version

题目描述

给你一个二叉树的根结点 root ，请返回出现次数最多的子树元素和。如果有多个元素出现的次数相同，返回所有出现次数最多的子树元素和（不限顺序）。

一个结点的 「子树元素和」 定义为以该结点为根的二叉树上所有结点的元素之和（包括结点本身）。

 

示例 1：

输入: root = [5,2,-3]
输出: [2,-3,4]

示例 2：

输入: root = [5,2,-5]
输出: [2]

 

提示:

• 节点数在 [1, 104] 范围内
• -105 <= Node.val <= 105

解法

方法一

# 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.
 * 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.
 * 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
  }
}