LeetCode_Convert Sorted List to Binary Search Tree-优快云博客

本文链接：https://blog.youkuaiyun.com/sheng_ai/article/details/46685605

一.题目

Convert Sorted List to Binary Search Tree

Total Accepted: 42911 Total Submissions: 153626 My Submissions

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

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二.解题技巧

这道题只要将列表转化为一个数组，就可以使用Convert Sorted Array to Binary Search Tree相同的方法来进行，这种方法的时间复杂度为O(n),空间复杂度为O(n^2)。

三.实现代码

#include <iostream>
#include <vector>

using std::vector;

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/


struct ListNode
{
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};


struct TreeNode
{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};


class Solution
{
private:
    TreeNode* sortedArrayToBST(vector<int>::iterator Begin,
                               vector<int>::iterator End)
    {
        if (Begin == End)
                {
                return NULL;
                }

        vector<int>::iterator HeadIte = Begin + (End - Begin) / 2;
        TreeNode *Head = new TreeNode(*HeadIte);

        TreeNode *LeftChild = sortedArrayToBST(Begin, HeadIte);
        TreeNode *RightChild = sortedArrayToBST(HeadIte + 1, End);

        Head->left = LeftChild;
        Head->right = RightChild;

        return Head;
    }

public:
    TreeNode* sortedListToBST(ListNode* head)
    {
        vector<int> Nums;

        for (ListNode *TmpHead = head; TmpHead; TmpHead = TmpHead->next)
        {
            Nums.push_back(TmpHead->val);
        }

        return sortedArrayToBST(Nums.begin(), Nums.end());
    }
};