Search Range in Binary Search Tree

Problem

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

If k1 = 10 and k2 = 22, then your function should return [12, 20, 22].

        20
       /  \
      8   22
     / \
    4   12

Code - Java

/**
 * 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> result = new ArrayList<Integer>();
        if (root == null) {
            return result;
        }
        if (root.val > k1) {
            result.addAll(searchRange(root.left, k1, k2));
        }
        if (root.val >= k1 && root.val <= k2) {
            result.add(root.val);
        }
        if (root.val < k2) {
            result.addAll(searchRange(root.right, k1, k2));
        }
        return result;
    }
}