LeetCode 508. Most Frequent Subtree Sum【一开始没看懂题】

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.

#include<iostream>#include<vector>#include<unordered_map>#include<queue>#include<map>using namespace std; struct TreeNode {     int val;     TreeNode *left;     TreeNode *right;     TreeNode(int x) : val(x), left(NULL), right(NULL) {} };class Solution {private:map<int, int> m;public:int singleNumber(vector<int>& nums) {//方法1/*unordered_map<int, int> m;for (int i = 0; i < nums.size(); i++){if (m.find(nums[i]) != m.end()){m[nums[i]]++;}else{m.insert({ nums[i], 1 });}}unordered_map<int, int> ::iterator itr = m.begin();while (itr != m.end()){if (itr->second != 2)return itr->first;itr++; }*///方法2int result = 0;for (int i = 0; i < nums.size(); i++){result = result ^ nums[i];}return result;}vector<int> largestValues(TreeNode* root) {//queue<int>q1;queue<TreeNode> q2;q2.push(*root);//q1.push(1);vector<int> ans;while (!q2.empty()){int size = q2.size();int maxval = (numeric_limits<int>::min)();for (int i = 0; i < size; i++){TreeNode treetmp = q2.front();if (treetmp.val > maxval)maxval = treetmp.val;  q2.pop();//q1.pop();if (treetmp.left != NULL){q2.push(*treetmp.left);}if (treetmp.right != NULL){q2.push(*treetmp.right);}}ans.push_back(maxval);}return ans;}vector<int> findFrequentTreeSum(TreeNode* root) {vector<int> ans;if (root == NULL)return ans;dfs(root);int maxn = 0;map<int, int>::iterator itr = m.begin();while (itr != m.end()){maxn = max(maxn, itr->second);itr++;}itr = m.begin();while (itr != m.end()){if (itr->second == maxn)ans.push_back(itr->first);itr++;}return ans;}void dfs(TreeNode* root) {if (root == NULL)return;if (root->left != NULL){dfs(root->left);root->val += root->left->val;}if (root->right != NULL){dfs(root->right);root->val += root->right->val;}m[root->val] ++;} };int main(){Solution s;//TreeNode n1(1);//TreeNode n2(3);//TreeNode n3(2);//TreeNode n4(5);//TreeNode n5(3);//TreeNode n6(9);//n1.left = &n2;//n1.right = &n3;//n2.left = &n4;//n2.right = &n5;//n3.right = &n6;//vector<int> v = s.largestValues(&n1);TreeNode n1(5);TreeNode n2(2);TreeNode n3(-5);n1.left = &n2;n1.right = &n3;vector<int> v = s.findFrequentTreeSum(&n1);return 0;}