Segment Tree Build

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本文介绍如何构建段树并实现区间查询最大值的方法。针对整数数组，通过递归方式构建段树，并设计查询方法返回指定区间的最大值。

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For an integer array (index from 0 to n-1, where n is the size of this array), in the corresponding SegmentTree, each node stores an extra attribute max to denote the maximum number in the interval of the array (index from start to end).

Design a query method with three parameters root, start and end, find the maximum number in the interval [start, end] by the given root of segment tree.

Notice

It is much easier to understand this problem if you finished Segment Tree Build first.

Have you met this question in a real interview?

Yes

Example

For array [1, 4, 2, 3], the corresponding Segment Tree is:

                  [0, 3, max=4]
                 /             \
          [0,1,max=4]        [2,3,max=3]
          /         \        /         \
   [0,0,max=1] [1,1,max=4] [2,2,max=2], [3,3,max=3]

query(root, 1, 1), return 4

query(root, 1, 2), return 4

query(root, 2, 3), return 3

query(root, 0, 2), return 4

这道题是要我们构建segmenttree，思路就是采用递归。

1. 首先写边界条件如果start>end 那么返回null

2. 如果 start == end那么说明当前位置为根结点，不用递归可以直接返回。

3. start < end，那么首先用start end构建一个当前节点，然后对它的左边，右边分别进行构建子数，然后再返回当前节点。

代码：

 public SegmentTreeNode build(int start, int end) {
        // write your code here
        if(start>end){
            return null;
        }
        if(start==end){
            return new SegmentTreeNode(start, start);
        }
        int mid = start + (end - start)/2;
        SegmentTreeNode node = new SegmentTreeNode(start, end);
        node.left = build(start, mid);
        node.right = build(mid+1, end);
        return node;
    }