11. Search Range in Binary Search Tree

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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转载自blog.csdn.net/foradawn/article/details/79948791