#201 Segment Tree Build

题目描述：

The structure of Segment Tree is a binary tree which each node has two attributesstart 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.

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 pointwhich 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] 题目思路：

这题就是简单的recursion。

Mycode（AC = 276ms）：

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