线段树的构造

http://www.lintcode.com/zh-cn/problem/segment-tree-build/

注意点：

1.  根节点用root表示
2.  把start == end的判断和后面代码合并，可以少一次return
3.  递归的两种条件： 逐层简化；最简单的情况下有返回值（终止条件）。所以在判断递归是否成立时，先考虑这两个条件是否已经满足。

 public SegmentTreeNode build(int start, int end) {
     if(start == end) return new SegmentTreeNode(start, end);
     if(start > end) return null;
     SegmentTreeNode st = new SegmentTreeNode(start, end);
     int left = start, right = (start + end) / 2;
     st.left = build(left, right);
     st.right = build(right + 1, end);
     return st; 
 }

public SegmentTreeNode build(int start, int end) {
        // write your code here
        if (start > end) {
            return null;
        }
        SegmentTreeNode root = new SegmentTreeNode(start, end);
        if(start != end) {
           int mid = (start + end) / 2;
           root.left = build(start, mid);
           root.right = build(mid + 1, end);
        }
        return root;
    }