Convert Sorted Array to Balanced BST in DSA Javascript - Execution Trace

Concept Flow - Convert Sorted Array to Balanced BST

Start with sorted array

↓

Find middle element

→Create root node with middle

↓

Left subarray

→Recursively build left subtree

↓

Right subarray

→Recursively build right subtree

↓

Attach left and right subtrees to root

↓

Return root of balanced BST

Split the array by middle element to create root, then recursively build left and right subtrees from subarrays.

Execution Sample

DSA Javascript

function sortedArrayToBST(arr) {
  if (!arr.length) return null;
  const mid = Math.floor(arr.length / 2);
  const root = { val: arr[mid] };
  root.left = sortedArrayToBST(arr.slice(0, mid));
  root.right = sortedArrayToBST(arr.slice(mid + 1));
  return root;
}

This code converts a sorted array into a balanced binary search tree by recursively choosing the middle element as root.

Execution Table

Step	Operation	Current Array	Middle Element	Node Created	Left Subtree Call	Right Subtree Call	Visual State
1	Start	[1,2,3,4,5,6,7]	4	Node(4)	[1,2,3]	[5,6,7]	4 / \
2	Left subtree root	[1,2,3]	2	Node(2)	[1]	[3]	4 / \ 2
3	Left subtree left leaf	[1]	1	Node(1)	[]	[]	4 / \ 2 /
4	Left subtree right leaf	[3]	3	Node(3)	[]	[]	4 / \ 2 3 /
5	Right subtree root	[5,6,7]	6	Node(6)	[5]	[7]	4 / \ 2 6 / \ / \
6	Right subtree left leaf	[5]	5	Node(5)	[]	[]	4 / \ 2 6 / \ / \ 1 3 5 7
7	Right subtree right leaf	[7]	7	Node(7)	[]	[]	4 / \ 2 6 / \ / \ 1 3 5 7
8	Return root	N/A	N/A	N/A	N/A	N/A	Balanced BST complete
9	Exit	N/A	N/A	N/A	N/A	N/A	All nodes created, recursion ends

💡 All subarrays processed, recursion ends with balanced BST root returned

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	After Step 7	Final
arr	[1,2,3,4,5,6,7]	[1,2,3,4,5,6,7]	[1,2,3]	[1]	[3]	[5,6,7]	[5]	[7]	N/A
mid	N/A	3	1	0	0	1	0	0	N/A
root.val	N/A	4	2	1	3	6	5	7	4
left subtree	N/A	pending	Node(2)	Node(1)	Node(3)	Complete	Complete	Complete	Complete
right subtree	N/A	pending	pending	pending	pending	Node(6)	Node(5)	Node(7)	Complete

Key Moments - 3 Insights

Why do we pick the middle element as root?

Picking the middle element (see Step 1 in execution_table) ensures the tree stays balanced by evenly splitting the array into left and right subtrees.

What happens when the subarray is empty?

When the subarray is empty (like in Step 3 and Step 4 calls with []), the function returns null, stopping recursion and creating leaf nodes.

How do left and right subtrees connect to the root?

After creating the root node (Step 1), recursive calls build left and right subtrees which are assigned to root.left and root.right respectively, as shown in Steps 2-7.

Visual Quiz - 3 Questions

Look at the execution_table, what is the middle element chosen at Step 5?

A 6

B 5

C 7

D 4

💡 Hint

Check the 'Middle Element' column at Step 5 in execution_table.

At which step does the recursion create the left leaf node with value 1?

A Step 2

B Step 3

C Step 4

D Step 6

💡 Hint

Look at the 'Node Created' column and 'Current Array' at Step 3.

If the input array was empty, what would the function return immediately?

A A node with value 0

B An error

C null

D An empty array

💡 Hint

Refer to the base case in the code sample where empty array returns null.

Concept Snapshot

Convert Sorted Array to Balanced BST:
- Pick middle element as root
- Recursively build left subtree from left half
- Recursively build right subtree from right half
- Base case: empty array returns null
- Result is height-balanced BST

Full Transcript

This concept converts a sorted array into a balanced binary search tree by choosing the middle element as the root node. The array is split into left and right halves recursively, building left and right subtrees. When the subarray is empty, recursion stops returning null, creating leaf nodes. This ensures the tree is balanced with minimal height. The code uses recursion and array slicing to build the tree step-by-step.