Convert Sorted Array to Binary Search Tree [LEETCODE]

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

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If you would have to choose an array element to be the root of a balanced BST, which element would you pick? The root of a balanced BST should be the middle element from the sorted array.

You would pick the middle element from the sorted array in each iteration. You then create a node in the tree initialized with this element. After the element is chosen, what is left? Could you identify the sub-problems within the problem?

There are two arrays left — The one on its left and the one on its right. These two arrays are the sub-problems of the original problem, since both of them are sorted. Furthermore, they are subtrees of the current node’s left and right child.

The code below creates a balanced BST from the sorted array in O(N) time (N is the number of elements in the array). Compare how similar the code is to a binary search algorithm. Both are using the divide and conquer methodology.

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode *sortedArrayToBST(vector<int> &num) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        return arrayToBST(num, 0, num.size());
    }
    TreeNode *arrayToBST(vector<int> &num, int start, int end) {
        if(end - start <= 0) {
            return NULL;
        }
        int mid = start + (end - start) / 2;
        TreeNode *root = new TreeNode(num[mid]);
        root->left = arrayToBST(num, start, mid);
        root->right = arrayToBST(num, mid + 1, end);
        return root;

    }
};