Search Range in Binary Search Tree

I put it here as a reminder to read question carefully. In this example, it asks to return result in ascending order, so that in-order traversal should be first consideration.

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.

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/** * 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) { // write your code here ArrayList<Integer> result = new ArrayList<Integer>(); if (root == null)return result; //it requires to return values in ascendiing order, so choose in order for that purpose inOrderTraversal(root,k1,k2,result); return result; } //using post order to stop left and right earlier void inOrderTraversal(TreeNode root,int k1, int k2,ArrayList<Integer> al){ if(root==null)return ; if(root.val>k1)inOrderTraversal(root.left,k1,k2,al); if(root.val>=k1 && root.val<=k2) al.add(root.val); if(root.val<k2)inOrderTraversal(root.right,k1,k2,al); return ; } }