Convert Sorted Array to Binary Search Tree


Problem Statement :


Given an integer array nums where the elements are sorted in ascending order, convert it to a 
height-balanced
 binary search tree.

 

Example 1:


Input: nums = [-10,-3,0,5,9]
Output: [0,-3,9,-10,null,5]
Explanation: [0,-10,5,null,-3,null,9] is also accepted:

Example 2:


Input: nums = [1,3]
Output: [3,1]
Explanation: [1,null,3] and [3,1] are both height-balanced BSTs.
 

Constraints:

1 <= nums.length <= 104
-104 <= nums[i] <= 104
nums is sorted in a strictly increasing order.


Solution :



title-img 


                            Solution in C :

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */


struct TreeNode* sortedArrayToBST(int* nums, int numsSize){
struct TreeNode* new_node;
if (numsSize==0) return NULL;
new_node = malloc(sizeof(struct TreeNode));
if (numsSize==1) {
    new_node->val = nums[0];
    new_node->right = NULL;
    new_node->left = NULL;
    return new_node;
}
new_node->left = sortedArrayToBST(nums,numsSize/2);
new_node->val = nums[numsSize/2];
new_node->right = sortedArrayToBST(nums+(numsSize/2)+1,numsSize-(1+(numsSize/2)));
return new_node;
}
                        


                        Solution in C++ :

class Solution {
public:
    TreeNode* sortedArrayToBST(vector<int>& nums) {
        return constructBSTRecursive(nums, 0, nums.size() - 1);
    }

    TreeNode* constructBSTRecursive(vector<int>& nums, int left, int right) {
        if(left > right)
            return NULL;
        int mid = left + (right - left) / 2;
        TreeNode* node = new TreeNode(nums[mid]);
        node->left = constructBSTRecursive(nums, left, mid - 1);
        node->right = constructBSTRecursive(nums, mid + 1, right);
        return node;
    }
};
                    


                        Solution in Java :

public class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        if(nums == null || nums.length == 0)
            return null;
        return constructBSTRecursive(nums, 0, nums.length - 1);
    }

    private TreeNode constructBSTRecursive(int[] nums, int left, int right) {
        if(left > right)
            return null;
        int mid = left + (right - left) / 2;
        TreeNode node = new TreeNode(nums[mid]);
        node.left = constructBSTRecursive(nums, left, mid - 1);
        node.right = constructBSTRecursive(nums, mid + 1, right);
        return node;
    }
}
                    


                        Solution in Python : 
                            
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]:
        if not nums: 
            return None 
        mid = len(nums) // 2 
        # create node and construct subtrees 
        node = TreeNode(nums[mid]) 
        node.left = self.sortedArrayToBST(nums[:mid]) 
        node.right = self.sortedArrayToBST(nums[mid+1:]) 
        return node


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