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.

Have you met this question in a real interview? 

Yes

Example

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

    20
   /  \
  8   22
 / \
4   12

/**
 * 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> res = new ArrayList<Integer>();
        
        Stack<TreeNode> stack = new Stack<TreeNode>();
        while(!stack.isEmpty() || root != null){
            if(root != null){
                if(root.val >= k1 && root.val <= k2){
                    stack.push(root);
                    root = root.left;
                }else if(root.val < k1){
                    root = root.right;
                } else if(root.val > k2){
                    root = root.left;
                }
            } else {
                TreeNode node = stack.pop();
                res.add(node.val);
                root = node.right;
            }
        }
        return res;
    }
}