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C++實(shí)現(xiàn)LeetCode(109.將有序鏈表轉(zhuǎn)為二叉搜索樹)

更新時(shí)間：2021年07月22日 15:09:26 作者：Grandyang

這篇文章主要介紹了C++實(shí)現(xiàn)LeetCode(109.將有序鏈表轉(zhuǎn)為二叉搜索樹),本篇文章通過簡(jiǎn)要的案例,講解了該項(xiàng)技術(shù)的了解與使用,以下就是詳細(xì)內(nèi)容,需要的朋友可以參考下

[LeetCode] 109.Convert Sorted List to Binary Search Tree 將有序鏈表轉(zhuǎn)為二叉搜索樹

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

這道題是要求把有序鏈表轉(zhuǎn)為二叉搜索樹，和之前那道 Convert Sorted Array to Binary Search Tree 思路完全一樣，只不過是操作的數(shù)據(jù)類型有所差別，一個(gè)是數(shù)組，一個(gè)是鏈表。數(shù)組方便就方便在可以通過index直接訪問任意一個(gè)元素，而鏈表不行。由于二分查找法每次需要找到中點(diǎn)，而鏈表的查找中間點(diǎn)可以通過快慢指針來操作，可參見之前的兩篇博客 Reorder List 和 Linked List Cycle II 有關(guān)快慢指針的應(yīng)用。找到中點(diǎn)后，要以中點(diǎn)的值建立一個(gè)數(shù)的根節(jié)點(diǎn)，然后需要把原鏈表斷開，分為前后兩個(gè)鏈表，都不能包含原中節(jié)點(diǎn)，然后再分別對(duì)這兩個(gè)鏈表遞歸調(diào)用原函數(shù)，分別連上左右子節(jié)點(diǎn)即可。代碼如下：

解法一：

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;
    }
};

我們也可以采用如下的遞歸方法，重寫一個(gè)遞歸函數(shù)，有兩個(gè)輸入?yún)?shù)，子鏈表的起點(diǎn)和終點(diǎn)，因?yàn)橹懒诉@兩個(gè)點(diǎn)，鏈表的范圍就可以確定了，而直接將中間部分轉(zhuǎn)換為二叉搜索樹即可，遞歸函數(shù)中的內(nèi)容跟上面解法中的極其相似，參見代碼如下：

解法二：

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;
    }
};

類似題目：

Convert Sorted Array to Binary Search Tree

參考資料：

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

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