Lintcode11 Search Range in Binary Search Tree solution 题解

【题目描述】

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.

给定两个值 k1 和 k2（k1 < k2）和一个二叉查找树的根节点。找到树中所有值在 k1 到 k2 范围内的节点。即打印所有x (k1 <= x <= k2) 其中 x 是二叉查找树的中的节点值。返回所有升序的节点值。

【题目链接】

http://www.lintcode.com/en/problem/search-range-in-binary-search-tree/

【题目解析】

中等偏易难度题，本题涉及到二叉查找树的按序输出，应马上联想到二叉树的中序遍历，对于二叉查找树而言，使用中序遍历即可得到有序元素。对每次访问的元素加以判断即可得最后结果，由于 OJ 上给的模板不适合递归处理，新建一个私有方法即可。

D & C。若root值比k1小，则只搜root右子树，若root值比k2大，则只搜root左子树，否则左右子树都搜，然后先加左子树返回的答案，再加root，最后加右子树返回的答案。

剪枝的判断条件容易出错，应将当前节点的值与k1和k2进行比较而不是其左子节点或右子节点的值。

【答案链接】

http://www.jiuzhang.com/solutions/search-range-in-binary-search-tree/

时间： 2024-10-16 11:58:46

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