Convert Sorted List to Binary Search Tree——将链表转换为平衡二叉搜索树 &&convert-sorted-array-to-binary-search-tree——将数列转换为bst

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.

为了满足平衡要求，容易想到提出中间节点作为树根，因为已排序，所以左右两侧天然满足BST的要求。

左右子串分别递归下去，上层根节点连接下层根节点即可完成。

递归找中点，然后断开前后两段链表，并继续找中点

 /**
  * 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) {
         if(head==NULL) return NULL;
         if(head->next==NULL) return new TreeNode(head->val);
         ListNode *mid=findMid(head);
         TreeNode *root=new TreeNode(mid->val);
         root->left=sortedListToBST(head);
         root->right=sortedListToBST(mid->next);
         return root;
     }
     ListNode *findMid(ListNode *root){
         if(root==NULL) return NULL;
         if(root->next==NULL) return root;
         ListNode *fast,*slow,*pre;
         fast=root;
         slow=root;
         while(fast!=NULL){
             fast=fast->next;
             if(fast!=NULL){
                 fast=fast->next;
                 pre=slow;
                 slow=slow->next;
             }
         }
         pre->next=NULL;
         return slow;
     }
 };

convert-sorted-array-to-binary-search-tree

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

 /**
  * 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) {
         int n=num.size();
         if(n<) return NULL;

         return findMid(num,,n-);
     }
     TreeNode *findMid(vector<int> &num, int l,int r){
         if(l>r) return NULL;
         int mid=(l+r+)/;
         TreeNode *root=new TreeNode(num[mid]);
         root->left=findMid(num,l,mid-);
         root->right=findMid(num,mid+,r);
         return root;
     }
 };