Lintcode: Search Range in Binary Search Tree

Given two values k1 and k2 (where k1 < k2) and a root pointer to a Binary Search Tree. Find all the keys of tree in range k1 to k2. i.e. print all x such that k1<=x<=k2 and x is a key of given BST. Return all the keys in ascending order.

Example
For example, if k1 = 10 and k2 = 22, then your function should print 12, 20 and 22.

          20

       /
    8           22

  /
4       12

我的做法是inorder traversal的变形，判断是否向左边递归的时候加上判断是否：root.val > k1, 如果否，则不需要继续向左递归；右子树的处理方法类似

 1 public class Solution {
 2     /**
 3      * @param root: The root of the binary search tree.
 4      * @param k1 and k2: range k1 to k2.
 5      * @return: Return all keys that k1<=key<=k2 in ascending order.
 6      */
 7     public ArrayList<Integer> searchRange(TreeNode root, int k1, int k2) {
 8         ArrayList<Integer> res = searchRangeRecur(root,k1,k2);
 9         return res;
10     }
11
12     public ArrayList<Integer> searchRangeRecur(TreeNode cur, int k1, int k2){
13         ArrayList<Integer> res = new ArrayList<Integer>();
14         if (cur==null) return res;
15         if (k1>k2) return res;
16
17         ArrayList<Integer> left = searchRangeRecur(cur.left,k1,Math.min(cur.val-1,k2));
18         ArrayList<Integer> right = searchRangeRecur(cur.right,Math.max(cur.val+1,k1),k2);
19
20         res.addAll(left);
21         if (cur.val>=k1 && cur.val<=k2) res.add(cur.val);
22         res.addAll(right);
23
24         return res;
25     }
26
27 }