LeetCode OJ：Convert Sorted Array to Binary Search Tree（将排序好的数组转换成二叉搜索树）...

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

讲一个排序好的数组转换成二叉搜索树，这题没想出来，基本上是参考别人的，边界条件应该注意一下：

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     TreeNode* sortedArrayToBST(vector<int>& nums) {
13         return createTree(nums, 0, nums.size() - 1);
14     }
15 
16     TreeNode * createTree(vector<int> & nums, int left, int right)
17     {
18         if(left > right)
19             return NULL;
20         int mid = left + (right - left)/2;
21         TreeNode * leftNode = createTree(nums, left, mid - 1);
22         TreeNode * rightNode = createTree(nums, mid + 1, right);
23         TreeNode * tmpRoot=  new TreeNode(nums[mid]);
24         tmpRoot->left = leftNode;
25         tmpRoot->right = rightNode;
26         return tmpRoot;
27     }
28 };

 java版本的代码如下所示，看到排好序的其实就应该注意一点了，自己肯定是不需要再进行比较插入的：

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public TreeNode sortedArrayToBST(int[] nums) {
12         return createTree(nums, 0, nums.length - 1);
13     }
14     
15     public TreeNode createTree(int[] nums, int left, int right){            
16         if(left > right)
17             return null;
18         int mid = left + (right-left) / 2;
19         TreeNode root = new TreeNode(nums[mid]);
20         root.left = createTree(nums, left, mid - 1);
21         root.right = createTree(nums, mid + 1, right);
22         return root;
23     }
24 }

 

转载于:https://www.cnblogs.com/-wang-cheng/p/4909637.html