[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 andend,
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.

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]

递归，按照描述计算left和right即可。

/**
* 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) {
return helper(start, end);
}

private SegmentTreeNode helper(int start, int end) {
if(start > end) return null;
SegmentTreeNode root = new SegmentTreeNode(start, end);
if(start == end) return root;

root.left = helper(start, (start + end) / 2);
root.right = helper((start + end) / 2 + 1, end);

return root;
}
}