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.

题意：给定一个有序的单向链表，将其转换成平衡的二叉搜索树。

思路：二分法+快慢指针

• 通过快慢指针找到根节点（链表中间节点）
• 左边部分用来构建左子树，右边部分用来构建右子树，然后递归链表左边部分和右边部分。

 

代码实现1：

class Solution {
public:
    // find middle element of the list
    ListNode *getmiddleList(ListNode *left,ListNode *right){
        //omit the condition :  left!=right && left->next!=right
        ListNode *pre,*last;
        pre=left; last =left->next;
        while(last!=right){
            last = last->next;
            if(last!=right){
                last = last->next;
                pre=pre->next;
            }
        }
        return pre;
    }
    
    // retri-BST constructor
    TreeNode *getBST(ListNode *left,ListNode *right){
        TreeNode *root = new TreeNode(0);
        //no leaf 
        if(left==right) return NULL;
        // only one leaf
        if(left->next == right){
            root->val=left->val;
            return root;
        }
        //more than one leaf
        ListNode *middle =getmiddleList(left,right);
        root->val = middle->val;
        root->left = getBST(left, middle);
        root->right = getBST(middle->next,right);
        return root;
    }
    TreeNode *sortedListToBST(ListNode *head) {
        TreeNode* root= new TreeNode(0);
        if(head==NULL) return NULL;
        if(head->next==NULL){
            root->val=head->val;
            root->left=root->right=NULL;
            return root;
        }
        ListNode *left,*middle,*right;
        middle=left=head;
        right=head->next;
        while(right){
            right=right->next;
            if(right){
                right=right->next;
                middle=middle->next;
            }
        }
        root->val=middle->val;
        root->left = getBST(left, middle);
        root->right= getBST(middle->next,right);
        return root;
    }
};     

 

代码实现2（不使用快慢指针）：

class Solution {
public:
    TreeNode* make(ListNode* head, int low, int high)
    {
        if(low > high)
            return NULL;

        int mid = (low + high) / 2;
        ListNode *p = head;
        for(int i=low; i <mid; i++)
        {
            p = p->next;
        }
        TreeNode *root = new TreeNode(p->val);
        root->left = make(head, low, mid-1);
        root->right = make(p->next, mid+1, high);
        return root;
    }

    TreeNode* sortedListToBST(ListNode* head) 
    {
        int n = 0;
        ListNode *p = head;
        while(p)
        {
            n++;
            p = p->next;
        }
        return make(head, 0, n-1);
    }
};