（树）根据排序数组或者排序链表重新构建BST树

• 题目一：给定一个数组，升序数组，将他构建成一个BST
• 思路：升序数组，这就类似于中序遍历二叉树得出的数组，那么根节点就是在数组中间位置，找到中间位置构建根节点，然后中间位置的左右两侧是根节点的左右子树，递归的对左右子树进行处理，得出一颗BST
• 代码：

  /**
   * 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) {
          return sortedArrayToBST(0, num.size()-1, num);
      }
      TreeNode *sortedArrayToBST(int left, int right, vector<int> &num){
          if (left > right)
              return NULL;
          int mid = (left + right)/2 + (left + right)%2 ;
          TreeNode *root = new TreeNode(num[mid]);
          root->left = sortedArrayToBST(left, mid-1, num);
          root->right = sortedArrayToBST(mid+1, right, num);
          return root;
      }
  };

   

• 题目二：和第一题类似，只不过数组变成了链表。
• 代码：

  /**
   * 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) {
          vector<int> num;
          if (head == NULL)
              return NULL;
          while (head){
              num.push_back(head->val);
              head = head->next;
          }
          return sortListToBST(0, num.size()-1, num);
      }
      TreeNode *sortListToBST(int start, int end, vector<int> &num){
          if (start > end)
              return NULL;
          int mid = (end + start)/2 + (end + start)%2;
          TreeNode *root = new TreeNode(num[mid]);
          root->left = sortListToBST(start, mid-1, num);
          root->right = sortListToBST(mid+1, end, num);
          return root;
      }
  };