[LeetCode] Convert Sorted List to Binary Search Tree

Convert Sorted List to Binary Search Tree

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

解题思路:

题意为构造有序链表的二分查找树。找到中间节点的办法用双指针法。注意我们还须要保存中间节点的前一个节点,便于一个链表分成两个链表。注意滴38行,是将链表恢复成原来的模样。

/**
 * 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;
        }
        ListNode* myHead = new ListNode(0);
        myHead->next = head;
        ListNode* pre = myHead;
        ListNode* one = head, *two = head;
        while(two->next!=NULL && two->next->next!=NULL){
            pre = pre->next;
            one = one->next;
            two = two->next->next;
        }
        two = one->next;
        pre->next = NULL;
        TreeNode* root = new TreeNode(one->val);
        root->left = sortedListToBST(myHead->next);
        root->right = sortedListToBST(two);
        pre->next = one;    //恢复链表结构
        //delete one; //这里为啥不能删除?
        delete myHead;
        return root;
    }
};


原文地址:https://www.cnblogs.com/mfmdaoyou/p/6784665.html