有序链表转化为平衡的二分查找树

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给定一个升序排列的有序单链表，将其转换为一棵平衡的二叉搜索树。

题目链接：https://oj.leetcode.com/problems/convert-sorted-list-to-binary-search-tree/

比较直观的解法是自顶向下的递归解决，先找到中间节点作为根节点，然后递归左右两部分。所有我们需要先找到中间节点，对于单链表来说，必须要遍历一边，可以使用快慢指针加快查找速度。

代码如下：

#include<iostream>#include<queue>using namespace std;/* node of binary tree */struct TreeNode{int data;TreeNode* lChild;TreeNode* rChild;TreeNode(int _data) { data = _data; lChild = rChild = nullptr; }};/* node of singly-linked list */struct ListNode{int data;ListNode* next;ListNode(int _data) { data = _data; next = nullptr; }};/* construct a singly-linked list */void ConstructList(ListNode* & _head){cout << "请输入一个升序的数组，以0结束： ";int d;ListNode* p;cin >> d;if (d){_head = new ListNode(d);p = _head;while (cin >> d&&d){p->next = new ListNode(d);p = p->next;}}}/* a sorted singly-linked list to a AVL*/TreeNode* SortedListToAVL(ListNode* & _head){if (!_head) return nullptr;if (!_head->next) return (new TreeNode(_head->data));//用快慢指针找到中间节点ListNode* preslow, *slow, *fast;preslow = nullptr;slow = fast = _head;while (fast->next&&fast->next->next){preslow = slow;slow = slow->next;fast = fast->next->next;}TreeNode* mid = new TreeNode(slow->data);//分别递归左右两部分if (preslow){preslow->next = nullptr;mid->lChild = SortedListToAVL(_head);}mid->rChild = SortedListToAVL(slow->next);return mid;}/* level order of binary tree */void LevelOrder(TreeNode* & _root){cout << "二叉树的层次遍历为: ";queue<TreeNode*> q;q.push(_root);while (!q.empty()){cout << q.front()->data << " ";if (q.front()->lChild)q.push(q.front()->lChild);if (q.front()->rChild)q.push(q.front()->rChild);q.pop();}cout << endl;}int main(){ListNode* listHead;TreeNode* TreeRoot;ConstructList(listHead);TreeRoot = SortedListToAVL(listHead);LevelOrder(TreeRoot);/*输入：1 2 3 4 5 6 7 8 9 0输出：5 2 7 1 3 6 8 4 9所构造的AVL是：                     5/ \   2   7  / \ / \ 1  3 6  8     \    \  4    9*/return 0;}

题目来自： http://www.acmerblog.com/convert-sorted-list-to-binary-search-tree-6124.html

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