树 - Convert Sorted List to Binary Search Tree

最新推荐文章于 2021-11-02 12:48:41 发布

一个不务正业的程序猿

最新推荐文章于 2021-11-02 12:48:41 发布

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本文链接：https://blog.youkuaiyun.com/cloud323/article/details/76604419

本文介绍了一种将升序排列的单链表转换为高度平衡二叉搜索树的方法。通过先将链表元素存储到数组中，再采用分而治之的策略递归构建平衡二叉树。

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一，Convert Sorted List to Binary Search Tree

题目描述

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

我的代码

/**
 * 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) {
        if(head == NULL){
            return NULL;
        }
        vector<int> nums;
        ListNode *p = head;
        while(p != NULL){
            nums.push_back(p->val);
            p = p->next;
        }
        return dfs(nums, 0, nums.size()-1);
    }
    
    TreeNode* dfs(vector<int>& nums, int s, int t){
        if(s <= t){
            int mid = (s + t) / 2;
            TreeNode *p = new TreeNode(nums[mid]);
            p->left = dfs(nums, s, mid-1);
            p->right = dfs(nums, mid+1, t);
            return p;
        } else {
            return NULL;
        }
    }
};