【Medium】201. Segment Tree Build

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.

Notice:

It's guaranteed that the size of the array is greater or equal to k.

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

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]

解题思路

Segment Tree(线段树) 是一种可以用来进行基于区间的操作的数据结构，所有满足结合律的操作都可以在Segment Tree上进行（比如元素出现次数、最大值、最小值etc.）。 创建、修改和查询Segment Tree的时间复杂度都是logn级别的，所以如果一道题能被转化成求在某个范围内满足某种条件，都可以用Segment Tree来做（比如找出比某个数大的，小的etc.）。

这题的follow up：

• 247. Segment Tree Query II

• 248 Count of Smaller Number

• 249 Count of Smaller Number before itself

核心代码

public SegmentTreeNode build(int start, int end) {
    // write your code here

    if (start > end){
        return null;
    }

    SegmentTreeNode node = new SegmentTreeNode(start, end);

    if (start == end){
        node.left = node.right = null;
    } else {
        int mid = start + (end - start) / 2;
        node.left = build(start, mid);
        node.right = build(mid + 1, end);
    }

    return node;
}

时间空间复杂度

O(n)

n为数组长度