LeetCode: Convert Sorted List to Binary Search Tree [109]

【题目】

Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

【题意】

将一个有序链表转换成平衡二叉树

【思路】

思路跟Convert Sorted Array to Binary Search Tree完全一样

【代码】

/** * 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:        ListNode *getMidNode(ListNode *head){        if(head==NULL || head->next==NULL)return head;        ListNode* prev=NULL;        //指向pstep1的前一个节点        ListNode* pstep1=head;      //每次向前移动一步        ListNode* pstep2=head->next;    //每次向前移动两步        //对于偶数链表，结束条件是pstep2->next==NULL        //对于计数链表，结束条件是pstep2==NULL                while(pstep2!=NULL && pstep2->next!=NULL){            prev=pstep1;            pstep1=pstep1->next;            pstep2=pstep2->next->next;        }                //找到链表中间节点的同时，把链表分裂为前后两个独立的链表        if(prev)prev->next=NULL;                return pstep1;    }    TreeNode *sortedListToBST(ListNode *head) {       if(head==NULL)return NULL;              ListNode*mid=getMidNode(head);       //创建根节点       TreeNode*root=(TreeNode*)malloc(sizeof(TreeNode));       root->val=mid->val;       //创建左子树       TreeNode*left=NULL;       if(mid!=head)            left=sortedListToBST(head);       //创建右子树       TreeNode*right=sortedListToBST(mid->next);       //构造二叉树       root->left=left;       root->right=right;              return root;           }};