<LeetCode OJ> 109 / 108 Convert Sorted ( List / Array ) to Binary Search Tree

Total Accepted: 71642 Total Submissions: 233074 Difficulty: Medium

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

分析：

有序单表链转化成二叉树，本题和有序数组转化成二叉搜索树差别还是有的，但是可以先将链表转化成数组就成为上一题的模式了！但是这样做恐怕没有达到考察目的！

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; *//** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode* sortedListToBST(ListNode* head) {        if( head == NULL )              return NULL;          vector<int> nums;        while(head)        {            nums.push_back(head->val);            head=head->next;        }        return creatTree(nums,0,nums.size()-1);    }        TreeNode* creatTree(vector<int>& nums,int leftpos,int rightpos)     {         if(leftpos > rightpos)              return NULL;         int midpos= (leftpos+rightpos)/2;            TreeNode* newnode=new TreeNode(nums[midpos]);         newnode->right=creatTree(nums,midpos+1,rightpos);         newnode->left=creatTree(nums,leftpos,midpos-1);         return newnode;     }  };

学习别人的算法：

快慢指针寻找链表的中间位置，原来此题就像考察这个！！！才明白！

class Solution {public:    TreeNode* sortedListToBST(ListNode* head) {        if(head==NULL)             return NULL;          if(head->next==NULL)             return new TreeNode(head->val);          return creatTree(head,NULL);    }        TreeNode* creatTree(ListNode* head,ListNode* tail)     {          if(head==tail)             return NULL; //head是起始节点，tail是终止结点的下一个结点          ListNode* slow=head;          ListNode* fast=head;          while(fast!=tail && fast->next!=tail)//通过快慢指针快速找到根节点          {              fast=fast->next->next;              slow=slow->next;          }          TreeNode* res=new TreeNode(slow->val);          res->left=creatTree(head,slow);//递归的构建左子树，右子树          res->right=creatTree(slow->next,tail);          return res;      }  };

Total Accepted: 77544 Total Submissions: 206040 Difficulty: Medium

Given an array where elements are sorted in ascending order, convert it to a height balanced BST.

分析：

有序数组转化成二叉搜索树。本题解题思路还是比较明显的！因为数组有序，为了能平衡，总是选择数组（未被使用过的位置的）中间位置的值作为当前二叉树的根节点！出门看了一下别人的做法，尽然不谋而合！

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode* sortedArrayToBST(vector<int>& nums) {        if( nums.size() == 0 )            return NULL;        return creatTree(nums,0,nums.size()-1);      }       TreeNode* creatTree(vector<int>& nums,int leftpos,int rightpos)   {       if(leftpos > rightpos)            return NULL;       int midpos= (leftpos+rightpos)/2;          TreeNode* newnode=new TreeNode(nums[midpos]);       newnode->right=creatTree(nums,midpos+1,rightpos);       newnode->left=creatTree(nums,leftpos,midpos-1);       return newnode;   }};

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原文地址：http://blog.csdn.net/ebowtang/article/details/51570438

原作者博客：http://blog.csdn.net/ebowtang

本博客LeetCode题解索引：http://blog.csdn.net/ebowtang/article/details/50668895