LeetCode: 109. Convert Sorted List to Binary Search Tree

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#leetcode #linkedlist #tree #binary tree #BST

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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。通过递归方法，文章详细阐述了创建平衡二叉树的过程，并提供了完整的AC代码实现。

LeetCode: 109. 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.

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:

题目大意: 根据给定的有序链表，生成一种可能的平衡二叉树。

解题思路

基本思路同 LeetCode: 108. Convert Sorted Array to Binary Search Tree 题解。由于给定数据结构是链表，因此只能先生成左子树，以使迭代指针指向链表中根节点的元素。

AC 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 {
private:
    // 将 head 前 n 个元素生成平衡二叉树 BST
    TreeNode* sortedListToBSTRef(ListNode*& head, int n)
    {
        if(head == nullptr || n <= 0) return nullptr;

        // 生成左子树
        TreeNode* leftChildTree = sortedListToBSTRef(head, n/2);

        // 生成根节点
        TreeNode* root = new TreeNode(head->val);
        head = head->next;

        // 生成右子树
        TreeNode* rightChildTree = sortedListToBSTRef(head, n - n/2 - 1);

        // 链接左/右子树
        root->left = leftChildTree;
        root->right = rightChildTree;

        return root;
    }
public:
    TreeNode* sortedListToBST(ListNode* head) {
        int n = 0; //记录节点个数

        // 计算节点个数
        ListNode* iter = head;
        while(iter != nullptr && ++n) iter = iter->next;

        return sortedListToBSTRef(head, n);
    }
};