（樹）根據排序陣列或者排序連結串列重新構建BST樹

• 題目一：給定一個陣列，升序陣列，將他構建成一個BST
• 思路：升序陣列，這就類似於中序遍歷二叉樹得出的陣列，那麼根節點就是在陣列中間位置，找到中間位置構建根節點，然後中間位置的左右兩側是根節點的左右子樹，遞迴的對左右子樹進行處理，得出一顆BST
• 程式碼：

  /**
   * 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;
      }
  };

   

• 題目二：和第一題類似，只不過陣列變成了連結串列。
• 程式碼：

  /**
   * 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;
      }
  };