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 a given array, so that we can create a corresponding segment tree with every node value represent the corresponding interval max value in the array, 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 [3,2,1,4]
. The segment tree will be:
[0, 3] (max = 4)
/ [0, 1] (max = 3) [2, 3] (max = 4)
/ \ / [0, 0](max = 3) [1, 1](max = 2)[2, 2](max = 1) [3, 3] (max = 4)
这道题是之前那道Segment Tree Build的拓展,这里面给线段树又增添了一个max变量,然后让我们用一个数组取初始化线段树,其中每个节点的max为该节点start和end代表的数组的坐标区域中的最大值。建树的方法跟之前那道没有什么区别,都是用递归来建立,不同的地方就是在于处理max的时候,如果start小于end,说明该节点还可以继续往下分为左右子节点,那么当前节点的max就是其左右子节点的max的较大值,如果start等于end,说明该节点已经不能继续分了,那么max赋为A[left]即可,参见代码如下:
class Solution { public: /** *@param A: a list of integer *@return: The root of Segment Tree */ SegmentTreeNode * build(vector<int>& A) { return build(A, 0, A.size() - 1); } SegmentTreeNode* build(vector<int>& A, int start, int end) { if (start > end) return NULL; SegmentTreeNode *node = new SegmentTreeNode(start, end); if (start < end) { node->left = build(A, start, (start + end) / 2); node->right = build(A, (start + end) / 2 + 1, end); node->max = max(node->left->max, node->right->max); } else { node->max = A[start]; } return node; } };
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