For an array, we can build aSegmentTreefor it, each node stores an extra attributecountto denote the number of elements in the the array which value is between interval start and end. (The array may not fully filled by elements)

Design aquerymethod with three parametersroot,startandend, find the number of elements in the in array's interval [start,end] by the given root of value SegmentTree.

Notice

It is much easier to understand this problem if you finishedSegment Tree BuildandSegment Tree Queryfirst.

Have you met this question in a real interview?

Yes

Example

For array[0, 2, 3], the corresponding value Segment Tree is:

                     [0, 3, count=3]
                     /             \
          [0,1,count=1]             [2,3,count=2]
          /         \               /            \
   [0,0,count=1] [1,1,count=0] [2,2,count=1], [3,3,count=1]

query(1, 1), return0

query(1, 2), return1

query(2, 3), return2

query(0, 2), return2

要注意有可能给出的范围大于当前节点的范围，直接return即可

/**
 * Definition of SegmentTreeNode:
 * public class SegmentTreeNode {
 *     public int start, end, count;
 *     public SegmentTreeNode left, right;
 *     public SegmentTreeNode(int start, int end, int count) {
 *         this.start = start;
 *         this.end = end;
 *         this.count = count;
 *         this.left = this.right = null;
 *     }
 * }
 */


public class Solution {
    /*
     * @param root: The root of segment tree.
     * @param start: start value.
     * @param end: end value.
     * @return: The count number in the interval [start, end]
     */
    public int query(SegmentTreeNode root, int start, int end) {
        // write your code here
        if (root == null) {
            return 0;
        }

        if (root.start >= start && root.end <= end) {
            return root.count;
        }

        int mid = (root.start + root.end) / 2;

        if (end <= mid) {
            return query(root.left, start, end);
        } else if (start > mid) {
            return query(root.right, start, end);
        }

        return query(root.left, start, mid) + query(root.right, mid + 1, end);
    }
}