Segment Tree Build

最新推荐文章于 2022-10-31 20:43:32 发布

flyatcmu

最新推荐文章于 2022-10-31 20:43:32 发布

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分类专栏： Segment Tree

本文链接：https://blog.youkuaiyun.com/u013325815/article/details/108810734

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本文详细介绍了一种名为段树的数据结构的构建过程。段树是一种二叉树，每个节点表示一个区间，通过递归方式创建左右子节点来划分区间，直至每个叶子节点表示单一元素。文章提供了具体的构建算法实现，并附带了示例。

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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.start, end=(A.start + A.end) / 2.
The right child of node A has start=(A.start + A.end) / 2 + 1, end=A.end.
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 start value, return the root of this segment tree.

Example

Example 1:

Input：[1,4]
Output："[1,4][1,2][3,4][1,1][2,2][3,3][4,4]"
Explanation：
	               [1,  4]
	             /        \
	      [1,  2]           [3, 4]
	      /     \           /     \
	   [1, 1]  [2, 2]     [3, 3]  [4, 4]

思路：基本模板，必须记住；

/**
 * 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: start value.
     * @param end: end value.
     * @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;
        }
        int mid = start + (end - start) / 2;
        node.left = build(start, mid);
        node.right = build(mid + 1, end);
        return node;
    }
}