Convert sorted list to binary search tree [leetcode] O (n) Algorithm

The main idea is similar to the middle-order traversal. First, we construct the left subtree, then the current node, and then the right subtree.

TreeNode *sortedListToBST(ListNode *head) {        int count = 0;        ListNode * cur = head;        while (cur)        {            count++;            cur = cur->next;        }        return sortedListToBST(head, count);    }        TreeNode *sortedListToBST(ListNode * (&head), int count) {        if (count <= 0) return NULL;        TreeNode* left = sortedListToBST(head, count / 2 );        TreeNode * root = new TreeNode(head->val);        head = head->next;        root->left = left;        root->right = sortedListToBST(head, count - (count / 2) - 1);        return root;    }

Another similar question: converting a binary search tree into a two-way linked list

Similar to the central traversal, the left subtree is converted into a two-way linked list first, and then the right subtree is processed.

The Code is as follows:

void BSTToList(TreeNode * t, ListNode * &l){if (t->left) BSTToList(t->left, l);ListNode * cur = new ListNode(t->val);cur->left = l;if (!l) l->right = cur;l = cur;if (t->right) BSTToList(t->right, l);}

Convert sorted list to binary search tree [leetcode] O (n) Algorithm