109. 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：其实这是个很经典的有序链表构建二叉查找树的问题，也就是先序遍历，后序遍历，中序遍历构建的问题，三者的代码如下：

1. 先序遍历：

class Solution {private:    TreeNode *helpsortedListToBST(ListNode *head, ListNode *tail){    if( head == tail )    return NULL;    ListNode *mid = head, *temp = head;    while( temp != tail && temp->next != tail ) {    mid = mid->next;    temp = temp->next->next;    }    TreeNode *root = new TreeNode(mid->val);    root->left = helpsortedListToBST(head, mid);    root->right = helpsortedListToBST(mid->next, tail );           return root;    }public:    TreeNode *sortedListToBST(ListNode *head){    return helpsortedListToBST( head, NULL );    }};

2.中序遍历：

class Solution {private:    TreeNode *helpsortedListToBST(ListNode *head, ListNode *tail){    if( head == tail )    return NULL;    ListNode *mid = head, *temp = head;    while( temp != tail && temp->next != tail ) {    mid = mid->next;    temp = temp->next->next;    }    TreeNode *root = new TreeNode(0);    root->left = helpsortedListToBST(head, mid);        root->val = mid->val;    root->right = helpsortedListToBST(mid->next, tail );    return root;    }public:    TreeNode *sortedListToBST(ListNode *head){    return helpsortedListToBST( head, NULL );    }};

3.后续遍历：

class Solution {private:    TreeNode *helpsortedListToBST(ListNode *head, ListNode *tail){    if( head == tail )    return NULL;    ListNode *mid = head, *temp = head;    while( temp != tail && temp->next != tail ) {    mid = mid->next;    temp = temp->next->next;    }    TreeNode *root = new TreeNode(0);    root->left = helpsortedListToBST(head, mid);    root->right = helpsortedListToBST(mid->next, tail );        root->val = mid->val;    return root;    }public:    TreeNode *sortedListToBST(ListNode *head){    return helpsortedListToBST( head, NULL );    }};

其实这三者的性能是一样的，因为其实先序，中序，后序，只不过是给当前根节点的赋值的顺序罢了，结果如下：

然后就是对上面的优化，每一个递归函数里面都有一个while，循环寻找中间节点，这是很耗时间的，我们想的是如何的去掉它，去掉他，也就是我们每次不去寻找中间节点，那么不是先序遍历，只能是中序遍历和后序遍历，但是因为中序遍历的的值正好是升序，左边->中间->右边，左边<中间<右边，而链表也是升序的，也就是说中序是最简单的，正好按链表顺序构建二叉查找树；

class Solution {    private:    ListNode* ls;    TreeNode *HelpsortedListToBST(int size){    if( size == 0 ) return NULL;        TreeNode *root = new TreeNode(0);    root->left = HelpsortedListToBST( size/2 );        root->val = ls->val;        ls = ls->next;    root->right = HelpsortedListToBST( size - size/2-1 );    return root;    }public:    TreeNode *sortedListToBST(ListNode *head){        int  size = 0;        ls = head;        while( head != NULL){            head=head->next;            size++;        }    return HelpsortedListToBST(size);    }};

对比一下，性能还是提升了13%的。