11. Search Range in Binary Search Tree

分手后的思念是犯贱 2022-05-27 08:39 194阅读 0赞

# 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;
        }
    }

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