[Leetcode] 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.

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方法一：借助Convert Sorted Array to Binary Search Tree 的方法，先将list转成vector即可，时间复杂度O（n），转化过程就是O（n），空间复杂度O（n）

class Solution {
    public:
        TreeNode *sortedListToBST(ListNode *head)
        {   
            vector<int> num;
            while(head != NULL)
            {   
                num.push_back(head->val);
                head = head->next;
            }   
            return sortedArrayToBST(num);
    
        }   
        TreeNode *sortedArrayToBST(vector<int> &num)
        {   
            int size = num.size();
            if(size == 0)
                return NULL;
             return sortedArrayToBSTInternal(num, 0, size - 1); 

        }   

        TreeNode *sortedArrayToBSTInternal(vector<int> &num, int low, int high)
        {   

            // the code is very important, i.e: low = 4, hight = 5, mid = 4, 
            // will call sortedArrayToBSTInternal(num, 4, 3)
            if(low > high)
                return NULL;
            if(low == high)
                return new TreeNode(num[low]);

            int mid = (high-low)/2 + low;
            cout << mid << endl;
            TreeNode *root = new TreeNode(num[mid]);
            TreeNode *left = sortedArrayToBSTInternal(num, low, mid - 1);
            TreeNode *right = sortedArrayToBSTInternal(num, mid + 1, high);
            root->left = left;
            root->right= right;
            return root;
        }
};

方法二：找出当前链表的中间节点，然后再递归左右的子链表，开始的时候程序先计算链表总厂，然后传入两个前后索引指针，最后每次递归找出中间节点即可。

时间复杂度O（n^2）,空间复杂度O（logn），递归导致stack用O（logn）

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int calLen(ListNode *node)
    {
        int len = 0;
        while(node)
        {
            len++;
            node = node->next;
        }
        return len;
    }
    
    TreeNode *createTree(ListNode *node, int left, int right)
    {
        if (left > right)
            return NULL;
            
        int mid = (left + right) / 2;
        
        ListNode *p = node;
        
        for(int i = left; i < mid; i++)
            p = p->next;
            
        TreeNode *leftNode = createTree(node, left, mid - 1);
        TreeNode *rightNode = createTree(p->next, mid + 1, right);
        
        TreeNode *tNode = new TreeNode(p->val);
        
        tNode->left = leftNode;
        tNode->right = rightNode;
        
        return tNode;        
    }
    
    TreeNode *sortedListToBST(ListNode *head) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int len = calLen(head);
        return createTree(head, 0, len - 1);
    }
};

方法3:

  bottom-up，时间复杂度 O(n)，空间复杂度 O(logn)

http://blog.csdn.net/salutlu/article/details/24502109  中递归

http://www.bwscitech.com/a/jishuzixun/javayuyan/2013/0930/15822.html

存在一种自底向上 (bottom-up) 的方法，见 http://leetcode.com/2010/11/convert-sorted-list-to-balanced-binary.html

class Solution {
    public:
        TreeNode *sortedListToBST(ListNode *head) {
            int len = 0;
            ListNode *p = head;
            while (p) {
                len++;
                p = p->next;
            }
            return sortedListToBST(head, 0, len - 1);
        }
    private:
        TreeNode* sortedListToBST(ListNode*& list, int start, int end) {
            if (start > end) return NULL;
            int mid = start + (end - start) / 2;
            TreeNode *leftChild = sortedListToBST(list, start, mid - 1);
            TreeNode *parent = new TreeNode(list->val);
            parent->left = leftChild;
            list = list->next;
            parent->right = sortedListToBST(list, mid + 1, end);
            return parent;
        }
};