Convert Sorted Array to Balanced BST in DSA C++ - 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 node

↓

Balanced BST constructed

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

Execution Sample

DSA C++

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

Recursively picks middle element as root and builds left and right subtrees from subarrays.

Execution Table

Step	Operation	Subarray Range	Node Created	Pointer Changes	Visual State
1	Find middle of [0..6]	[0..6]	Node(4)	root = Node(4)	4
2	Build left subtree of 4	[0..2]	Node(2)	root->left = Node(2)	4 / 2
3	Build left subtree of 2	[0..0]	Node(1)	Node(2)->left = Node(1)	4 / 2 / 1
4	Build left subtree of 1	[0..-1]	None	Node(1)->left = nullptr	No change
5	Build right subtree of 1	[1..0]	None	Node(1)->right = nullptr	No change
6	Build right subtree of 2	[2..2]	Node(3)	Node(2)->right = Node(3)	4 / 2 / \ 1 3
7	Build left subtree of 3	[2..1]	None	Node(3)->left = nullptr	No change
8	Build right subtree of 3	[3..2]	None	Node(3)->right = nullptr	No change
9	Build right subtree of 4	[4..6]	Node(6)	root->right = Node(6)	4 / \ 2 6 / \
10	Build left subtree of 6	[4..4]	Node(5)	Node(6)->left = Node(5)	4 / \ 2 6 / \ / 1 3 5
11	Build left subtree of 5	[4..3]	None	Node(5)->left = nullptr	No change
12	Build right subtree of 5	[5..4]	None	Node(5)->right = nullptr	No change
13	Build right subtree of 6	[6..6]	Node(7)	Node(6)->right = Node(7)	4 / \ 2 6 / \ / \ 1 3 5 7
14	Build left subtree of 7	[6..5]	None	Node(7)->left = nullptr	No change
15	Build right subtree of 7	[7..6]	None	Node(7)->right = nullptr	No change
16	Return root node	N/A	N/A	N/A	Balanced BST complete

💡 Recursion ends when left > right, no nodes created for empty subarrays.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 6	After Step 9	After Step 13	Final
left	0	0	0	0	0	4	6	N/A
right	6	6	2	0	2	6	6	N/A
mid	N/A	3	1	0	2	5	6	N/A
root	nullptr	Node(4)	Node(4)	Node(4)	Node(4)	Node(4)	Node(4)	Node(4)
current node	N/A	Node(4)	Node(2)	Node(1)	Node(3)	Node(6)	Node(7)	N/A

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 is balanced because left and right subarrays have nearly equal sizes.

What happens when the subarray is empty?

When left > right (e.g., Step 4), recursion returns nullptr, meaning no node is created, stopping further branching.

How do left and right pointers get assigned?

After creating a node, recursive calls assign left and right children (see Steps 2, 6, 9, 13) by attaching returned nodes to pointers.

Visual Quiz - 3 Questions

Look at the execution_table at Step 6, which node is created and attached?

A Node(3) attached as right child of Node(2)

B Node(5) attached as left child of Node(6)

C Node(1) attached as left child of Node(2)

D Node(7) attached as right child of Node(6)

💡 Hint

Check Step 6 row in execution_table under 'Node Created' and 'Pointer Changes'

At which step does the recursion stop for the left subtree of Node(1)?

A Step 5

B Step 4

C Step 7

D Step 8

💡 Hint

Look for subarray range where left > right and no node created in execution_table

If the input array had 5 elements instead of 7, how would the root node change?

A Root would be the last element

B Root would be the first element

C Root would be the middle element of the 5 elements

D Root would be the average of all elements

💡 Hint

Recall the concept_flow and Step 1 where middle element is chosen as root

Concept Snapshot

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

Full Transcript

This concept converts a sorted array into a balanced binary search tree by recursively selecting the middle element as the root node. The left half of the array forms the left subtree, and the right half forms the right subtree. The recursion stops when the subarray is empty. Each step creates nodes and attaches them as left or right children, ensuring the tree remains balanced. The execution table traces each recursive call, node creation, and pointer assignment, showing the tree's growth visually. Key moments clarify why the middle element is chosen, how recursion ends, and how pointers are assigned. The visual quiz tests understanding of node creation and recursion stopping points. The snapshot summarizes the process in simple steps.