链表转平衡二叉搜索树-优快云博客

本文介绍如何将一个升序排列的单链表转换为高度平衡的二叉搜索树（BST）。通过将链表元素复制到数组中，利用二分查找确定子树根节点并递归构建BST。

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Convert Sorted List to Binary Search Tree

问题描述：

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

解题思路：

先将链表复制到一个数组里，再对数组使用二分查找作为子树根，递归构造二分查找树即可

代码：

/**
 * 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) {}
 * };
 */
class Solution {
public:
    TreeNode* sortedListToBST(ListNode* head) {
        vector<int> nums(0);
        ListNode *p = head;
        while(p != NULL){
            nums.push_back(p -> val);
            p = p -> next;
        }
        TreeNode* root = NULL;
        BSTreeCons(nums, 0, nums.size() - 1, root);
        return root;
    }
        void BSTreeCons(vector<int>& nums, int begin, int end, TreeNode*& root){
        if(begin > end) return;
        int mid = (begin + end) / 2;
        root = new TreeNode(nums[mid]);
        BSTreeCons(nums, begin, mid - 1, root -> left);
        BSTreeCons(nums, mid + 1, end, root -> right);
        
    }
};

结果：