LeetCode-探索-初级算法-树-5. 将有序数组转换为二叉搜索树(个人做题记录，不是习题讲解)

LeetCode探索-初级算法：https://leetcode-cn.com/explore/interview/card/top-interview-questions-easy/

5. 将有序数组转换为二叉搜索树

• 语言：java

• 思路：感觉和上一题的层层递归的有点像，再加上左右子树最多高度相差一，我就想说每次取数组中值当作当前树的根节点，左边当作左子树，右边当作右子树，有点二分查找的感觉。没想到一次过了，挺高兴的。

• 代码(1ms)：

  /**
   * Definition for a binary tree node.
   * public class TreeNode {
   *     int val;
   *     TreeNode left;
   *     TreeNode right;
   *     TreeNode(int x) { val = x; }
   * }
   */
  class Solution {
      public TreeNode sortedArrayToBST(int[] nums) {
          return BST(nums);
      }
      
      public TreeNode BST(int[] nums){
          int len = nums.length;
          if(len>0){
              int middle = len/2;
              TreeNode node = new TreeNode(nums[middle]);
              node.left = BST(Arrays.copyOfRange(nums,0,middle));
              node.right = BST(Arrays.copyOfRange(nums,middle+1,len));
              return node;
          }
          return null;
      }
  }
  

• 参考代码(1ms):感觉和我的差别不大，只不过它改用下标来代替获取子数组的方式，而我是直接暴力获取子数组(如果测试数据很大，肯定参考代码的优势就明显了)

  class Solution {
      public TreeNode sortedArrayToBST(int[] nums) {
          return toBST(nums, 0, nums.length - 1);
      }
      public TreeNode toBST(int[] nums, int i, int j) {
          if (i > j) {
              return null;
          }
          int mid = i + (j - i) / 2;
          TreeNode node = new TreeNode(nums[mid]);
          node.left = toBST(nums, i, mid - 1);
          node.right = toBST(nums, mid + 1, j);
          return node;
      }
  }
  

• 参考后重写：

  /**
   * Definition for a binary tree node.
   * public class TreeNode {
   *     int val;
   *     TreeNode left;
   *     TreeNode right;
   *     TreeNode(int x) { val = x; }
   * }
   */
  class Solution {
      public TreeNode sortedArrayToBST(int[] nums) {
          return BST(nums,0,nums.length-1);
      }
      
      public TreeNode BST(int[] nums,int start,int end){
          if(end>=start){
              int middle = start + (end-start)/2;
              TreeNode node = new TreeNode(nums[middle]);
              node.left = BST(nums,start,middle-1);
              node.right = BST(nums,middle+1,end);
              return node;
          }
          return null;
      }
  }