#439 Segment Tree Build II

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 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 pointwhich 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) 题目思路：

这题先从最下面的子树找起，start==end的时候，max就是A[start]。找完子树的max就可以得到root的max了。还是用recursion就行。

Mycode（AC = 86ms）：

/** * Definition of SegmentTreeNode: * class SegmentTreeNode { * public: * int start, end, max; * SegmentTreeNode *left, *right; * SegmentTreeNode(int start, int end, int max) { * this->start = start; * this->end = end; * this->max = max; * this->left = this->right = NULL; * } * } */ class Solution { public: /** *@param A: a list of integer *@return: The root of Segment Tree */ SegmentTreeNode * build(vector<int>& A) { // write your code here return buildTree(A, 0, A.size() - 1); } SegmentTreeNode* buildTree(vector<int>& A, int start, int end) { if (start > end) { return NULL; } else if (start == end) { return new SegmentTreeNode(start, end, A[start]); } else { SegmentTreeNode *root = new SegmentTreeNode(start, end, A[start]); root->left = buildTree(A, start, (start + end) / 2); root->right = buildTree(A, (start + end) / 2 + 1, end); int max_val = INT_MIN; if (root->left) max_val = max(max_val, root->left->max); if (root->right) max_val = max(max_val, root->right->max); root->max = max_val; return root; } } };