LintCode-Search Range in Binary Search Tree

LiBlog發表於2015-01-01

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

Solution:

 1 /**
 2  * Definition of TreeNode:
 3  * public class TreeNode {
 4  *     public int val;
 5  *     public TreeNode left, right;
 6  *     public TreeNode(int val) {
 7  *         this.val = val;
 8  *         this.left = this.right = null;
 9  *     }
10  * }
11  */
12 public class Solution {
13     /**
14      * @param root: The root of the binary search tree.
15      * @param k1 and k2: range k1 to k2.
16      * @return: Return all keys that k1<=key<=k2 in ascending order.
17      */
18     public ArrayList<Integer> searchRange(TreeNode root, int k1, int k2) {
19         ArrayList<Integer> res = searchRangeRecur(root,k1,k2);
20         return res;
21     }
22 
23     public ArrayList<Integer> searchRangeRecur(TreeNode cur, int k1, int k2){
24         ArrayList<Integer> res = new ArrayList<Integer>();
25         if (cur==null) return res;
26         if (k1>k2) return res;
27 
28         ArrayList<Integer> left = searchRangeRecur(cur.left,k1,Math.min(cur.val-1,k2));
29         ArrayList<Integer> right = searchRangeRecur(cur.right,Math.max(cur.val+1,k1),k2);
30 
31         res.addAll(left);
32         if (cur.val>=k1 && cur.val<=k2) res.add(cur.val);
33         res.addAll(right);
34 
35         return res;
36     }
37         
38 }

 

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