[LintCode] Segment Tree Build 建立线段树

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 startand 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.

Have you met this question in a real interview?

Yes

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]

这道题让我们建立线段树，也叫区间树，是一种高级树结构，但是题目中讲的很清楚，所以这道题实现起来并不难，我们可以用递归来建立，写法很简单，参见代码如下：

class Solution {
public:
/**
*@param start, end: Denote an segment / interval
*@return: The root of Segment Tree
*/
SegmentTreeNode * build(int start, int end) {
if (start > end) return NULL;
SegmentTreeNode *node = new SegmentTreeNode(start, end);
if (start < end) {
node->left = build(start, (start + end) / 2);
node->right = build((start + end) / 2 + 1, end);
}
return node;
}
};