|
| 1 | +--- |
| 2 | +id: convert-sorted-array-to-binary-search-tree |
| 3 | +title: Convert Sorted Array to Binary Search Tree |
| 4 | +sidebar_label: 0108-convert-sorted-array-to-binary-search-tree |
| 5 | +tags: |
| 6 | + - Array |
| 7 | + - Binary search tree |
| 8 | + - Divide and Conquer |
| 9 | + - Tree |
| 10 | + - Binary Tree |
| 11 | +description: "Given an integer array nums where the elements are sorted in ascending order, convert it to a height-balanced binary search tree." |
| 12 | +--- |
| 13 | + |
| 14 | + |
| 15 | +### Problem Description |
| 16 | + |
| 17 | +Given an integer array nums where the elements are sorted in ascending order, convert it to a height-balanced binary search tree. |
| 18 | + |
| 19 | +### Examples |
| 20 | + |
| 21 | +#### Example 1 |
| 22 | + |
| 23 | +- **Input:** ` nums = [-10,-3,0,5,9]` |
| 24 | +- **Output:** `[0,-3,9,-10,null,5]` |
| 25 | + |
| 26 | +#### Example 2 |
| 27 | + |
| 28 | +- **Input:** ` nums = [1,3]` |
| 29 | +- **Output:** `[3,1]` |
| 30 | + |
| 31 | +### Constraints |
| 32 | + |
| 33 | +- $1 \leq \text{nums.length} \leq 10^4$ |
| 34 | +- $-10^4 \leq \text{nums}[i] \leq 10^4$ |
| 35 | +- nums is sorted in a strictly increasing order |
| 36 | + |
| 37 | +### Solution Code |
| 38 | + |
| 39 | +#### Python |
| 40 | + |
| 41 | +```python |
| 42 | +class Solution: |
| 43 | + def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]: |
| 44 | + total_nums = len(nums) |
| 45 | + if not total_nums: |
| 46 | + return None |
| 47 | + |
| 48 | + mid_node = total_nums // 2 |
| 49 | + return TreeNode( |
| 50 | + nums[mid_node], |
| 51 | + self.sortedArrayToBST(nums[:mid_node]), self.sortedArrayToBST(nums[mid_node + 1 :]) |
| 52 | + ) |
| 53 | +``` |
| 54 | + |
| 55 | +#### Java |
| 56 | + |
| 57 | +```java |
| 58 | +class Solution { |
| 59 | + public TreeNode sortedArrayToBST(int[] nums) { |
| 60 | + int n=nums.length; |
| 61 | + if (n ==0) return null; |
| 62 | + if (n ==1) { |
| 63 | + TreeNode head = new TreeNode(nums[0]); |
| 64 | + return head; |
| 65 | + } |
| 66 | + |
| 67 | + TreeNode head = buildBST(nums,0,n-1); |
| 68 | + return head; |
| 69 | + } |
| 70 | + private TreeNode buildBST (int [] nums, int low, int high){ |
| 71 | + if(low>high) return null; |
| 72 | + |
| 73 | + int mid =low+(high-low)/2; |
| 74 | + TreeNode node = new TreeNode(nums[mid]); |
| 75 | + node.left =buildBST(nums,low,mid-1); |
| 76 | + node.right =buildBST(nums,mid+1,high); |
| 77 | + return node; |
| 78 | + } |
| 79 | +} |
| 80 | +``` |
| 81 | + |
| 82 | +#### C++ |
| 83 | + |
| 84 | +```cpp |
| 85 | +class Solution { |
| 86 | +public: |
| 87 | + TreeNode* sortedArrayToBST(vector<int>& nums) { |
| 88 | + return constructBSTRecursive(nums, 0, nums.size() - 1); |
| 89 | + } |
| 90 | + |
| 91 | + TreeNode* constructBSTRecursive(vector<int>& nums, int left, int right) { |
| 92 | + if(left > right) |
| 93 | + return NULL; |
| 94 | + int mid = left + (right - left) / 2; |
| 95 | + TreeNode* node = new TreeNode(nums[mid]); |
| 96 | + node->left = constructBSTRecursive(nums, left, mid - 1); |
| 97 | + node->right = constructBSTRecursive(nums, mid + 1, right); |
| 98 | + return node; |
| 99 | + } |
| 100 | +}; |
| 101 | +``` |
| 102 | +#### Javascript |
| 103 | + |
| 104 | +```javascript |
| 105 | +var sortedArrayToBST = function(nums) { |
| 106 | + if(!nums.length) return null; |
| 107 | + let mid = Math.floor(nums.length / 2); |
| 108 | + let root = new TreeNode(nums[mid]); |
| 109 | + root.left = sortedArrayToBST(nums.slice(0, mid)); |
| 110 | + root.right = sortedArrayToBST(nums.slice(mid + 1)); |
| 111 | + return root; |
| 112 | +}; |
| 113 | +``` |
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