LeetCode解题报告（367) -- 109. Convert Sorted List to Binary Search Tree

Problem

Given the head of 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 1:

1
2
3

Input: head = [-10,-3,0,5,9]
Output: [0,-3,9,-10,null,5]
Explanation: One possible answer is [0,-3,9,-10,null,5], which represents the shown height balanced BST.

Example 2:

1 2	Input: head = [] Output: []

Example 3:

1 2	Input: head = [0] Output: [0]

Example 4:

1 2	Input: head = [1,3] Output: [3,1]

Constraints:

The number of nodes in head is in the range [0, 2 * 104].
-10^5 <= Node.val <= 10^5

Analysis

题目要求根据一个有序链表去构建一颗BST。我们知道BST中序遍历的结果就是有序的，所以可以通过这个特点反过来构建BST。这里我们先不考虑数据结构，假设用的是数组，怎么构建？

首先根节点是最中间的元素，然后左手边的元素就是左子树，右手边的元素就是右子树，只需要递归下去构建即可。这样看来，如果用的是数组，直接指定下标即可，但是这里给的是链表，稍微复杂一点。我们每次递归时，都需要找出中间的节点，方法是用快慢指针，找到中间的节点后，左边的元素扔进去左子树递归，节点右边的元素扔进去右子树递归即可。

Solution

无。

Code

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
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;
    }
};

Summary

这道题目本质上还是由数组构建BST的变种，这里换成了链表，整体的思路是不变的，唯一变化的是需要用快慢指针找中间节点。这道题目的分享到这里，感谢你的支持！