Search Range in Binary Search Tree

最新推荐文章于 2022-01-20 12:12:47 发布

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最新推荐文章于 2022-01-20 12:12:47 发布

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原文链接：http://www.cnblogs.com/beiyeqingteng/p/5642227.html

本文介绍了一种在二叉搜索树中查找指定区间内所有元素的方法。通过递归遍历，当根节点值小于区间下限时遍历右子树；当根节点值大于区间上限时遍历左子树；当根节点值位于区间内时，收集该值并分别遍历左右子树。

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.

Have you met this question in a real interview?

Yes

Example

If k1 = 10 and k2 = 22, then your function should return[12, 20, 22].

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分析：

这题还是很简单的，只要根据root的值和k1, k2的值分三种情况讨论就可以了。

 1 /**
 2  * Definition of TreeNode:
 3  * public class TreeNode {
 4  *     public int val;
 5  *     public TreeNode left, right;
 6  *     public TreeNode(int val) {
 7  *         this.val = val;
 8  *         this.left = this.right = null;
 9  *     }
10  * }
11  */
12 public class Solution {
13     /**
14      * @param root: The root of the binary search tree.
15      * @param k1 and k2: range k1 to k2.
16      * @return: Return all keys that k1<=key<=k2 in ascending order.
17      */
18     public ArrayList<Integer> searchRange(TreeNode root, int k1, int k2) {
19         // write your code here
20         ArrayList<Integer> list = new ArrayList<Integer>();
21         if (k1 <= k2) {
22             traverse(root, k1, k2, list);
23         }
24         Collections.sort(list);
25         return list;
26     }
27     
28     public void traverse(TreeNode node, int k1, int k2, ArrayList<Integer> list) {
29         if (node != null) {
30             if (node.val < k1) {
31                 traverse(node.right, k1, k2, list);
32             } else if (node.val > k2) {
33                 traverse(node.left, k1, k2, list);
34             } else {
35                 list.add(node.val);
36                 traverse(node.left, k1, k2, list);
37                 traverse(node.right, k1, k2, list);
38             }
39         }
40     }
41 }