[Lintcode]Segment Tree Build

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本文详细介绍了一种名为区间树的数据结构的构建过程。通过递归方式创建区间树，并为每个节点分配正确的开始和结束值。提供了具体的实现代码示例，帮助读者理解如何从给定的开始和结束值构建出完整的区间树。

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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 hasstart=A.left, end=(A.left + A.right) / 2.
The right child of node A hasstart=(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;
    }
}