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.ExampleFor 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 }