leetcode 508. Most Frequent Subtree Sum 子树和 + 一个简单的DFS深度优先遍历的做法

不念不忘少年蓝@ 2022-06-02 22:48 125阅读 0赞

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.

本题题意很简单，就是考察每一个子树subtree的统计情况

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <unordered_map>
    #include <set>
    #include <unordered_set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    #include <regex>
    #include <iomanip>
    
    using namespace std;
    
    
    #include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <cmath>
    
    using namespace std;
    
    
    
    /*
    struct TreeNode
    {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
    };
    */
    
    class Solution
    {
    public:
        int maxCount = 0;
        map<int, int> mmp;
        vector<int> findFrequentTreeSum(TreeNode* root)
        {
            vector<int> res;
            getAll(root);
            for (auto a : mmp)
            {
                if (a.second == maxCount)
                    res.push_back(a.first);
            }
            return res;
        }
    
        int getAll(TreeNode* root)
        {
            if (root == NULL)
                return 0;
            else
            {
                int sum = root->val + getAll(root->left) + getAll(root->right);
                mmp[sum] += 1;
                maxCount = max(maxCount, mmp[sum]);
                return sum;
            }
        }
    };