Count of Smaller Numbers After Self

You are given an integer array nums and you have to return a new counts array.The counts array has the property where counts[i] is the number of smaller elements to the right ofnums[i].

Example:

Given nums = [5, 2, 6, 1]To the right of 5 there are 2 smaller elements (2 and 1).To the right of 2 there is only 1 smaller element (1).To the right of 6 there is 1 smaller element (1).To the right of 1 there is 0 smaller element.

Return the array [2, 1, 1, 0].

class Solution {struct node{    int val,copy,leftCnt;    node *left,*right;    node(int x){val=x;copy=1;leftCnt=0;left=NULL;right=NULL;}};public:    vector<int> countSmaller(vector<int>& nums) {        int curSum;        int sz = nums.size();        vector<int> r(sz,0);        if(sz <= 1) return r;         node *root = new node(nums[sz-1]);        node *t;        node *pos;        int v;        for (int i=sz-2;i>=0;i--){            curSum = 0;            t = root;            v = nums[i];                        while(t){                pos = t;                if(v < t->val){                    t->leftCnt++;                    t = t->left;                }                 else if(v == t->val){                    t->copy++;                    curSum += t->leftCnt;                    break;                }                else {                    curSum += (t->leftCnt + t->copy);                    t = t->right;                }            }                        if(v < pos->val){                pos->left = new node(v);            }            else if(v > pos->val){                pos->right = new node(v);            }                        r[i] = curSum;        }        return r;    }};

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