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;

}
};