11. Search Range in Binary Search Tree-蒲公英云

11. Search Range in Binary Search Tree

Description

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

Solution

import java.util.*;
/** * 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 { 
    List<Integer> list = new ArrayList<>();
    /** * @param root: param root: The root of the binary search tree * @param k1: An integer * @param k2: An integer * @return: return: Return all keys that k1<=key<=k2 in ascending order */
    public List<Integer> searchRange(TreeNode root, int k1, int k2) {
        // write your code here
        if(k1>k2) return list;
        if(root!=null){
            searchRange(root.left,k1,k2);
            if(root.val>=k1 && root.val<=k2) list.add(root.val);
            searchRange(root.right,k1,k2);
        }
        return list;
    }
}