LeetCode:Convert Sorted Array to Binary Search Tree,Convert Sorted List to Binary Search Tree

LeetCode:Convert Sorted Array to Binary Search Tree

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

分析：找到数组的中间数据作为根节点，小于中间数据的数组来构造作为左子树，大于中间数据的数组来构造右子树，递归解法如下

 /**
  * Definition for binary tree
  * struct TreeNode {
  * int val;
  * TreeNode *left;
  * TreeNode *right;
  * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  * };
  */
 class Solution {
 public:
     TreeNode *sortedArrayToBST(vector<int> &num) {
         // IMPORTANT: Please reset any member data you declared, as
         // the same Solution instance will be reused for each test case.
         int len = num.size();
         if(len == )return NULL;
         return sortedArrayToBSTRecur(num, , len-);
     }
     TreeNode *sortedArrayToBSTRecur(vector<int> &num, int istart, int iend)
     {
         if(istart > iend)return NULL;
         int middle = (istart+iend)/;
         TreeNode *res = new TreeNode(num[middle]);
         res->left = sortedArrayToBSTRecur(num, istart, middle-);
         res->right = sortedArrayToBSTRecur(num, middle+, iend);
         return res;
     }
 };

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LeetCode:Convert Sorted List to Binary Search Tree

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

分析：和上一题同理，只不过要使用快慢指针来找到链表的中间节点                                                  本文地址

 /**
  * 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:
     TreeNode *sortedListToBST(ListNode *head) {
         // IMPORTANT: Please reset any member data you declared, as
         // the same Solution instance will be reused for each test case.
         if(head == NULL)return NULL;
         ListNode *fast = head, *slow = head, *preSlow = NULL;
         while(fast->next && fast->next->next)
         {
             fast = fast->next->next;
             preSlow = slow;
             slow = slow->next;
         }
         TreeNode *res = new TreeNode(slow->val);
         fast = slow->next;
         delete slow;
         if(preSlow != NULL)
         {
             preSlow->next = NULL;
             res->left = sortedListToBST(head);
         }
         res->right = sortedListToBST(fast);
         return res;
     }
 };

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