Interval Minimum Number(区间最小数)

http://www.lintcode.com/zh-cn/problem/interval-minimum-number/
请参阅
Segment Tree Build II(线段树的构造 II)
Segment Tree Query II(线段树查询 II)
Segment Tree Modify(线段树的修改)

import java.util.ArrayList;

public class Solution {

    /**
     * @param A, queries: Given an integer array and an query list
     * @return: The result list
     */
    public ArrayList intervalMinNumber(int[] A,List queries) {
        // write your code here
        Node node = build(A, 0, A.length - 1);
        ArrayList list = new ArrayList<>();
        for (Interval temp : queries) {
            list.add(query(node, temp.start, temp.end));
        }
        return list;
    }

    private int query(Node root, int start, int end) {
        if (end < start) {
            return 0;
        }
        if (root.start == start && root.end == end) {
            return root.min;
        }
        int mid = (root.start + root.end) >>> 1;
        if (start >= mid + 1) {
            return query(root.right, start, end);
        } else if (end <= mid) {
            return query(root.left, start, end);
        } else {
            return Math.min(query(root.left, start, mid), query(root.right, mid + 1, end));
        }
    }


    private Node build(int[] a, int l, int r) {
        Node node = new Node(l, r, a[l]);
        if (l == r) {
            return node;
        }
        int mid = (l + r) >>> 1;
        node.left = build(a, l, mid);
        node.right = build(a, mid + 1, r);
        node.min = Math.min(node.left.min, node.right.min);
        return node;
    }

    private class Node {
        int start, end;
        int min;
        Node left, right;

        public Node(int start, int end, int min) {
            this.start = start;
            this.end = end;
            this.min = min;
        }
    }
}