Convert Sorted List to Binary Search Tree [leetcode] O(n)的算法

主要的思想类似中序遍历，先构建左子树，再构建当前节点，并构建右子树

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;    }

还有一个类似的题目：将二叉搜索树转换成双向链表

同样是类似中序遍历，先将左子树变成双向链表，再处理右子树

代码如下：

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);}