*[Lintcode]Segment Tree Build II

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 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.

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)

正向递归从上向下计算区间，用回溯算法从下向上计算区间内最大值，避免在区间内进行搜索。

/** * Definition of SegmentTreeNode: * public class SegmentTreeNode { *     public int start, end, max; *     public SegmentTreeNode left, right; *     public SegmentTreeNode(int start, int end, int max) { *         this.start = start; *         this.end = end; *         this.max = max *         this.left = this.right = null; *     } * } */public class Solution {    /**     *@param A: a list of integer     *@return: The root of Segment Tree     */    public SegmentTreeNode build(int[] A) {        return helper(A, 0, A.length - 1);    }        private SegmentTreeNode helper(int[] A, int start, int end) {        if(start > end) return null;                SegmentTreeNode root = new SegmentTreeNode(start, end);        if(start == end) {            root.max = A[start];            return root;        }                root.left = helper(A, start, (start + end) / 2);        root.right = helper(A, (start + end) / 2 + 1, end);                //back tracking. Avoiding unnecessaary number.        root.max = Math.max(root.left.max, root.right.max);        return root;    }}