LeetCode -- Count of Smaller Numbers After Self

public class Solution {    public IList<int> CountSmaller(int[] nums)     {        if (nums == null || nums.Length == 0){    		return new List<int>();    	}    	    	var result = new List<int>(){0 /*there is no value on the right of last number*/};    	var len = nums.Length;    	var root = new BTNode(nums[len - 1]);    	for (var i = len - 2; i >= 0; i--){    		// add the value of arr on the tree one by one    		// also update the count of smaller on the 'root' node    		var smallerCount = BuildAndCount(root, nums[i]);    		result.Add(smallerCount);    	}    	    	result.Reverse();    	    	return result;     }          public int BuildAndCount(BTNode current, int value){         	int countOfSmaller = 0;     	while(true){    		if(value > current.Value){    			    			// value bigger than current node , add current node's 'count of smaller' on this value    			countOfSmaller += current.Count + 1;    			// keep going right ,until reach leaf    			if(current.Right == null){    				current.Right = new BTNode(value);    				break;    			}else{    				current = current.Right;    			}    		}    		else{    			// value is smaller than current node value , increase the count of smaller on current node    			current.Count ++;    			// put value on left tree leaf    			if(current.Left == null){    				current.Left = new BTNode(value);    				break;    			}    			else{    				current = current.Left;    			}    		}    	}    	    	return countOfSmaller;     }          public class BTNode{     	public BTNode Left;    	public BTNode Right;    	public int Value;    	public int Count ;    	public BTNode (int v){    		Value = v;    		Count = 0;    	}     }}也可以考虑BIT的实现：https://www.topcoder.com/community/data-science/data-science-tutorials/binary-indexed-trees/