# Segment Tree Build

## Problem

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.

## 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

## Code - Java

/**
* 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) {
if (start > end) {
return null;
}
SegmentTreeNode node = new SegmentTreeNode(start, end);
if (start == end) {
return node;
}

node.left = build(start, (start + end) / 2);
node.right = build((start + end) / 2 + 1, end);

return node;
}
}