Get Keys In Binary Search Tree In Given Range - Easy

Posted 2021-02-01 fatttcat

Get the list of keys in a given binary search tree in a given range[min, max] in ascending order, both min and max are inclusive.

Examples

        5

      /    

    3        8

  /          

 1     4        11

get the keys in [2, 5] in ascending order, result is  [3, 4, 5]

Corner Cases

• What if there are no keys in the given range? Return an empty list in this case.

 

 

inorder traversal + 剪枝

如果当前node.val > min，走左子树；如果当前node.val在min~max之间，加入res中；如果当前node.val < max，走右子树

time: O(h + # of nodes in [min, max]), space: O(h)

/**
 * public class TreeNode {
 *   public int key;
 *   public TreeNode left;
 *   public TreeNode right;
 *   public TreeNode(int key) {
 *     this.key = key;
 *   }
 * }
 */
public class Solution {
  public List<Integer> getRange(TreeNode root, int min, int max) {
    // Write your solution here
    List<Integer> res = new ArrayList<>();
    if(root == null || min > max) {
      return res;
    }
    helper(root, min, max, res);
    return res;
  }
  
  public void helper(TreeNode root, int min, int max, List<Integer> res) {
    if(root == null) {
      return;
    }
    if(root.key > min) {
      helper(root.left, min, max, res);
    }
    if(root.key >= min && root.key <= max) {
      res.add(root.key);
    }
    if(root.key < max){
      helper(root.right, min, max, res);
    }
  }
}