Convert Sorted Array to Binary Search Tree LeetCode Solution

Problem – Convert Sorted Array to Binary Search Tree

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

A height-balanced binary tree is a binary tree in which the depth of the two subtrees of every node never differs by more than one.

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.

Convert Sorted Array to Binary Search Tree LeetCode Solution in Java

public TreeNode sortedArrayToBST(int[] num) {
    if (num.length == 0) {
        return null;
    }
    TreeNode head = helper(num, 0, num.length - 1);
    return head;
}

public TreeNode helper(int[] num, int low, int high) {
    if (low > high) { // Done
        return null;
    }
    int mid = (low + high) / 2;
    TreeNode node = new TreeNode(num[mid]);
    node.left = helper(num, low, mid - 1);
    node.right = helper(num, mid + 1, high);
    return node;
}

Convert Sorted Array to Binary Search Tree LeetCode Solution in Python

# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param num, a list of integers
    # @return a tree node
    # 12:37
    def sortedArrayToBST(self, num):
        if not num:
            return None

        mid = len(num) // 2

        root = TreeNode(num[mid])
        root.left = self.sortedArrayToBST(num[:mid])
        root.right = self.sortedArrayToBST(num[mid+1:])

        return root

Convert Sorted Array to Binary Search Tree LeetCode Solution in C++

class Solution {
public:
    TreeNode *sortedArrayToBST(vector<int> &num) {
        if(num.size() == 0) return NULL;
        if(num.size() == 1)
        {
            return new TreeNode(num[0]);
        }
        
        int middle = num.size()/2;
        TreeNode* root = new TreeNode(num[middle]);
        
        vector<int> leftInts(num.begin(), num.begin()+middle);
        vector<int> rightInts(num.begin()+middle+1, num.end());
        
        root->left = sortedArrayToBST(leftInts);
        root->right = sortedArrayToBST(rightInts);
        
        return root;
    }
};

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