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

Solution:

/**
* Definition of TreeNode:
* public class TreeNode {
*     public int val;
*     public TreeNode left, right;
*     public TreeNode(int val) {
*         this.val = val;
*         this.left = this.right = null;
*     }
* }
*/
public class Solution {
/**
* @param root: The root of the binary search tree.
* @param k1 and k2: range k1 to k2.
* @return: Return all keys that k1<=key<=k2 in ascending order.
*/
public ArrayList<Integer> searchRange(TreeNode root, int k1, int k2) {
ArrayList<Integer> res = searchRangeRecur(root,k1,k2);
return res;
}

public ArrayList<Integer> searchRangeRecur(TreeNode cur, int k1, int k2){
ArrayList<Integer> res = new ArrayList<Integer>();
if (cur==null) return res;
if (k1>k2) return res;

ArrayList<Integer> left = searchRangeRecur(cur.left,k1,Math.min(cur.val-1,k2));
ArrayList<Integer> right = searchRangeRecur(cur.right,Math.max(cur.val+1,k1),k2);

res.addAll(left);
if (cur.val>=k1 && cur.val<=k2) res.add(cur.val);
res.addAll(right);

return res;
}

}