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 }