[LintCode] Segment Tree Build 建立线段树-白红宇

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 hasstart=A.left, end=(A.left + A.right) / 2.

The right child of node A hasstart=(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:

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;    }};

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