Leetcode 315. 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 of 

nums[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]

.

从后到前依次插入节点，每个节点储存 比它小的节点数 和 当前节点的值。

public class TreeNode {
int cnt;
int val;
TreeNode left;
TreeNode right;
public TreeNode(int cnt, int val) {
this.cnt = cnt;
this.val = val;
}
}

public List<Integer> countSmaller(int[] nums) {
TreeNode root = null;
Integer[] res = new Integer[nums.length];
if (nums == null || nums.length == 0) return Arrays.asList(res);
int len = nums.length;
for (int i = len - 1; i >= 0; i--) {
root = insert(root, nums[i], res, i, 0);
}
return Arrays.asList(res);
}

private TreeNode insert(TreeNode root, int val, Integer[] ans, int idx, int pre) {
if (root == null) {
root = new TreeNode(0, val);
ans[idx] = pre;
}
else if (root.val > val) {
root.cnt++;
root.left = insert(root.left, val, ans, idx, pre);
}
else root.right = insert(root.right, val, ans, idx, pre + root.cnt + (root.val < val ? 1 : 0));
return root;
}