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：

有序單錶鏈轉化成二叉樹，本題和有序數組轉化成二叉搜索樹差別還是有的，但是可以先將鏈表轉化成數組就成爲上一題的模式了！但是這樣做恐怕沒有達到考察目的！

/**
 * 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;  
   }  
};

學習別人的算法：

快慢指針尋找鏈表的中間位置，原來此題就像考察這個！！！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是起始節點，tail是終止結點的下一個結點  
        ListNode* slow=head;  
        ListNode* fast=head;  
        while(fast!=tail && fast->next!=tail)//通過快慢指針快速找到根節點  
        {  
            fast=fast->next->next;  
            slow=slow->next;  
        }  
        TreeNode* res=new TreeNode(slow->val);  
        res->left=creatTree(head,slow);//遞歸的構建左子樹，右子樹  
        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：

有序數組轉化成二叉搜索樹。本題解題思路還是比較明顯的！因爲數組有序，爲了能平衡，總是選擇數組（未被使用過的位置的）中間位置的值作爲當前二叉樹的根節點！出門看了一下別人的做法，盡然不謀而合！

/**
 * 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;
   }
};

注：本博文爲EbowTang原創，後續可能繼續更新本文。如果轉載，請務必複製本條信息！

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題解索引：http://blog.csdn.net/ebowtang/article/details/50668895