2265. number of nodes with statistical value equal to the average value of subtree

2265. The number of nodes whose statistical value is equal to the average value of the subtree \

Give you the root node of a binary tree root , Find and return the number of nodes that meet the requirements , The value of the node is required to be equal to its subtree median Average .

Be careful ：

n  The average value of elements can be determined by  n  Elements   Sum up   Then divide by  n , and   Round down   To the nearest integer .
root  Of   subtree   from  root  And all its descendants .

Example 1：

Input ：root = [4,8,5,0,1,null,6]
Output ：5
explain ：
The right value is 4 The node of ： Average value of subtree (4 + 8 + 5 + 0 + 1 + 6) / 6 = 24 / 6 = 4 .
The right value is 5 The node of ： Average value of subtree (5 + 6) / 2 = 11 / 2 = 5 .
The right value is 0 The node of ： Average value of subtree 0 / 1 = 0 .
The right value is 1 The node of ： Average value of subtree 1 / 1 = 1 .
The right value is 6 The node of ： Average value of subtree 6 / 1 = 6 .

/** * Definition for a binary tree node. * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * }; */
 int num;
 int* dfs(struct TreeNode* root,int *data){
    
     if(root){
    
         int *a=dfs(root->left,data);
         int numa=a[0];
         int vala=a[1];
         int *b=dfs(root->right,data);
         int numb=b[0];
         int valb=b[1];
         int sum=vala+valb+root->val;
         if(sum/(numa+numb+1)==root->val){
    
             num++;
         }
         data[0]=numa+numb+1;
         data[1]=sum;
         return data;
     }
     else{
    
                data[0]=0;
                data[1]=0;
         return data;
     }

 }
int averageOfSubtree(struct TreeNode* root){
    
    num=0;
     int *data=(int*)malloc(sizeof(int)*2);
    int *a=dfs(root,data);
    return num;

}