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

1. 如果root.val > k1 递归的找他的左子树

2. 如果root.val 在k1, k2之间 添加root到result

3. 如果root.val < k2 递归找他的右子树

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>();
        helper(root, k1, k2, result);
        return result;
    }
    private void helper(TreeNode root, int k1, int k2, ArrayList<Integer> result){
        if (root == null){
            return;
        }
        if (root.val > k1){
            helper(root.left, k1, k2, result);
        }
        if (root.val >= k1 && root.val <= k2){
            result.add(root.val);
        }
        if (root. val < k2){
            helper(root.right, k1, k2, result);
        }
    }
}