[LintCode] 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

Note

重点是：根据BST的性质，先左后根最后右。
另一重点是，helper函数和main函数都要用的的参数，记得在main函数外层定义。

Solution

public class Solution {
    private ArrayList<Integer> result;
    public ArrayList<Integer> searchRange(TreeNode root, int k1, int k2) {
        // write your code here
        result = new ArrayList<Integer>();
        Searchfunc(root, k1, k2);
        return result;
    }
    public void Searchfunc(TreeNode root, int k1, int k2) {
        if (root == null) {
            return;
        }
        if (root.val > k1) {
            Searchfunc(root.left, k1, k2);
            }
        if (root.val >= k1 && root.val <= k2) {
            result.add(root.val);
        }
        if (root.val < k2) {
            Searchfunc(root.right, k1, k2);
        }
    }
}