109. Convert Sorted List to Binary Search Tree

Q

https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/?tab=Description
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

A

法一

思想同数组convert sorted array to binary search tree，只是链表只能顺序遍历。

 struct TreeNode* toBST(struct ListNode* head, int n){     if (n == 0) {         return NULL;     }     struct ListNode *p;     struct TreeNode *node;     int mid;     int i;     p = head;     mid = n/2;     for (i = 0; i < mid; ++i) {         p = p->next;     }     node = (struct TreeNode *)malloc(sizeof(struct TreeNode));     node->val = p->val;     node->left = toBST(head, mid);     p = p->next;     node->right = toBST(p, n-mid-1);     return node; }struct TreeNode* sortedListToBST(struct ListNode* head) {    if (head == NULL) {        return NULL;    }    int n = 0;    struct ListNode* p;    p = head;    while(p != NULL) {        ++n;        p = p->next;    }    return toBST(head, n);}

方法二，模拟树的中序遍历，自底向上建树，注意这里指向指针的指针

C语言

struct TreeNode *toBST(struct ListNode **head, int n) {    if(n == 0) {        return NULL;    }    /*    int mid;    struct TreeNode *node;    node = (struct TreeNode *)malloc(sizeof(struct TreeNode));    node->left = toBST(head, mid);    node->val = head->val;    head = head->next;    node->right = toBST(head, n-mid-1);    return node;    */    int mid;    struct TreeNode *c, *p;    //struct ListNode *h;    mid = n/2;    c = toBST(head, mid);    p = (struct TreeNode *)malloc(sizeof(struct TreeNode));    p->val = (*head)->val;    p->left = c;    *head = (*head)->next;    p->right = toBST(head, n-mid-1);    return p;}struct TreeNode* sortedListToBST(struct ListNode* head) {    if (head == NULL) {        return NULL;    }    int n = 0;    struct ListNode* p;    p = head;    while(p != NULL) {        ++n;        p = p->next;    }    return toBST(&head, n);}