<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.

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Depth-first Search Linked
List

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(M) Convert Sorted Array to Binary Search Tree

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.

Subscribe to see which companies asked this question

Hide Tags
Tree Depth-first
Search

Hide Similar Problems
(M) Convert Sorted List to Binary Search Tree

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;
}
};

注：本博文为EbowTang原创，后续可能继续更新本文。如果转载，请务必复制本条信息！

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题解索引：/article/3664871.html