108. Convert Sorted Array to Binary Search Tree (Divide and Conquer) 20.3.18 Easy

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

For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of 
every node never differ by more than 1.

Example:

Given the sorted array: [-10,-3,0,5,9],

One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:

      0
     / \
   -3   9
   /   /
 -10  5
 
 Solution: -------------------------------------------------- Two Pointers + DFS
 -------------------------------------- O(N) T, O(logN) S
 # Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def sortedArrayToBST(self, nums):
        return self.to_bst_help(nums, 0, len(nums)-1)
        """
        :type nums: List[int]
        :rtype: TreeNode                                     # It requires us to return a node
        """
    def to_bst_help(self, nums, start, end):
        if start > end:                                      # This condition used very often, need to be remembered
            return None                                      # "return None" or "return" are all OK !!!!!!!!!!!!!!!!!!!!!!!!
        mid = (start + end) / 2                              # '/' in python 2 is '//'
                                                             # Even though Python does not overflow,
                                                             # it is still better to change mid = (start + end) / 2
                                                             # into mid = start + (end - start) / 2 and explain it to the interviewers
        node = TreeNode(nums[mid])                           # Can define a node using its build-in class
        node.left = self.to_bst_help(nums, start, mid - 1)   # Since the function need to return a node, we should use assignment here
        node.right = self.to_bst_help(nums, mid + 1, end)     
        '''
        node.left = nums[mid - 1]                            # Why cannot do like this?
                                                             # mid may be zero  e.x.: (0 + 1)/ 2 = 0
                                                             # and node.left should be None in the end as in the line 34
                                                             # So, use the recursion function for assignment
                                                             # THINK LIKE A PROGRAMMER!
        node.right = nums[end]
        self.to_bst_help(nums, start, mid - 1)
        self.to_bst_help(nums, mid + 1, end)           
        '''
        return node                                          # As the question required 



Java Version:
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode sortedArrayToBST(int[] nums) {
        if (nums == null || nums.length == 0) {
            return null;
        }
        return build(nums, 0, nums.length - 1);
    }
    
    public TreeNode build(int[] nums, int left, int right) {
        if (left > right) {
            return null;
        }
        if (left == right) {
            return new TreeNode(nums[left]);
        }
        int mid = left + (right - left) / 2;
        TreeNode currNode = new TreeNode(nums[mid]);
        currNode.left = build(nums, left, mid - 1);
        currNode.right = build(nums, mid + 1, right);
        return currNode;
    }
}