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C++实现LeetCode(109.将有序链表转为二叉搜索树)

时间:2021-12-02 14:17     来源/作者:Grandyang

[LeetCode] 109.Convert Sorted List to Binary Search Tree 将有序链表转为二叉搜索树

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example:

Given the sorted linked list: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
/ \
-3   9
/   /
-10  5

这道题是要求把有序链表转为二叉搜索树,和之前那道 Convert Sorted Array to Binary Search Tree 思路完全一样,只不过是操作的数据类型有所差别,一个是数组,一个是链表。数组方便就方便在可以通过index直接访问任意一个元素,而链表不行。由于二分查找法每次需要找到中点,而链表的查找中间点可以通过快慢指针来操作,可参见之前的两篇博客 Reorder List 和 Linked List Cycle II 有关快慢指针的应用。找到中点后,要以中点的值建立一个数的根节点,然后需要把原链表断开,分为前后两个链表,都不能包含原中节点,然后再分别对这两个链表递归调用原函数,分别连上左右子节点即可。代码如下:

解法一:

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class Solution {
public:
    TreeNode *sortedListToBST(ListNode* head) {
        if (!head) return NULL;
        if (!head->next) return new TreeNode(head->val);
        ListNode *slow = head, *fast = head, *last = slow;
        while (fast->next && fast->next->next) {
            last = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        fast = slow->next;
        last->next = NULL;
        TreeNode *cur = new TreeNode(slow->val);
        if (head != slow) cur->left = sortedListToBST(head);
        cur->right = sortedListToBST(fast);
        return cur;
    }
};

我们也可以采用如下的递归方法,重写一个递归函数,有两个输入参数,子链表的起点和终点,因为知道了这两个点,链表的范围就可以确定了,而直接将中间部分转换为二叉搜索树即可,递归函数中的内容跟上面解法中的极其相似,参见代码如下:

解法二:

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class Solution {
public:
    TreeNode* sortedListToBST(ListNode* head) {
        if (!head) return NULL;
        return helper(head, NULL);
    }
    TreeNode* helper(ListNode* head, ListNode* tail) {
        if (head == tail) return NULL;
        ListNode *slow = head, *fast = head;
        while (fast != tail && fast->next != tail) {
            slow = slow->next;
            fast = fast->next->next;
        }
        TreeNode *cur = new TreeNode(slow->val);
        cur->left = helper(head, slow);
        cur->right = helper(slow->next, tail);
        return cur;
    }
};

类似题目:

Convert Sorted Array to Binary Search Tree

参考资料:

https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/

https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/discuss/35476/Share-my-JAVA-solution-1ms-very-short-and-concise.

https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/discuss/35470/Recursive-BST-construction-using-slow-fast-traversal-on-linked-list

到此这篇关于C++实现LeetCode(109.将有序链表转为二叉搜索树)的文章就介绍到这了,更多相关C++实现将有序链表转为二叉搜索树内容请搜索服务器之家以前的文章或继续浏览下面的相关文章希望大家以后多多支持服务器之家!

原文链接:https://www.cnblogs.com/grandyang/p/4295618.html

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