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
buildmethod. - 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) / );
node->right = build((start + end) / + , end);
}
return node;
}
};