Lintcode: Segment Tree Build

The structure of Segment Tree is a binary tree which each node has two attributes start and end denote an segment / interval.

start and end are both integers, they should be assigned in following rules:

The root's start and end is given by build method.
The left child of node A has start=A.left, end=(A.left + A.right) / 2.
The right child of node A has start=(A.left + A.right) / 2 + 1, end=A.right.
if start equals to end, there will be no children for this node.
Implement a build method with two parameters start and end, so that we can create a corresponding segment tree with every node has the correct start and end value, return the root of this segment tree.

Have you met this question in a real interview? Yes
Example
Given start=0, end=3. The segment tree will be:

[0,  3]
/        \
[0,  1]           [2, 3]
/     \           /     \
[0, 0]  [1, 1]     [2, 2]  [3, 3]
Given start=1, end=6. The segment tree will be:

[1,  6]
/        \
[1,  3]           [4,  6]
/     \           /     \
[1, 2]  [3,3]     [4, 5]   [6,6]
/    \           /     \
[1,1]   [2,2]     [4,4]   [5,5]
Clarification
Segment Tree (a.k.a Interval Tree) is an advanced data structure which can support queries like:

which of these intervals contain a given point
which of these points are in a given interval
See wiki:
Segment Tree
Interval Tree

/**
* Definition of SegmentTreeNode:
* public class SegmentTreeNode {
*     public int start, end;
*     public SegmentTreeNode left, right;
*     public SegmentTreeNode(int start, int end) {
*         this.start = start, this.end = end;
*         this.left = this.right = null;
*     }
* }
*/
public class Solution {
/**
*@param start, end: Denote an segment / interval
*@return: The root of Segment Tree
*/
public SegmentTreeNode build(int start, int end) {
// write your code here
if (start > end) return null;
if (start == end) return new SegmentTreeNode(start, start);
SegmentTreeNode cur = new SegmentTreeNode(start, end);
cur.left = build(cur.start, (cur.start+cur.end)/2);
cur.right = build((cur.start+cur.end)/2+1, cur.end);
return cur;
}
}