Segment Tree Build 201
Question
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.
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
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]
Solution
用d&c的方法建立左右子树后连起来即可。
代码如下:
/**
* 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, end);
}
int mid = (start + end) / 2;
SegmentTreeNode root = new SegmentTreeNode(start, end);
SegmentTreeNode left = build(start, mid);
SegmentTreeNode right = build(mid + 1, end);
root.left = left;
root.right = right;
return root;
}
}