Balance a Binary Search Tree - Algorithm, Visualization, Examples

Contents

ProblemExamplesSolutionAlgorithmCode

Problem Statement❯

You are given an unbalanced Binary Search Tree (BST). Your task is to transform it into a height-balanced BST. A height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differs by more than 1.

Your solution should not change the values of the nodes, only the structure of the tree to ensure it becomes balanced.

Examples❯

Input Tree Balanced BST Output Description
[1, null, 2, null, null, null, 3]
[2, 1, 3]
Original tree is right-skewed and unbalanced. In-order traversal: [1,2,3] → balanced BST is [2,1,3].
[3, 2, null, 1]
[2, 1, 3]
Input tree is left-heavy. In-order traversal: [1,2,3] → balanced BST root is 2 with 1 and 3 as children.
[10, 5, 15, 2, 7, null, 20]
[10, 5, 15, 2, 7, null, 20]
Already balanced BST. In-order traversal is sorted and tree height is balanced → output is unchanged.
[] [] Empty tree is trivially balanced; no action needed.
[7]
[7]
Single-node tree is inherently balanced; remains unchanged.

Solution❯

To balance a Binary Search Tree (BST), the idea is to rebuild it in a way that the height difference between left and right subtrees is minimal. We do this by leveraging the inorder traversal property of BSTs, which gives us the sorted list of elements.

Case 1: Tree with Multiple Nodes (e.g., [1, 2, 3, 4, 5])

In this case, the BST might be skewed — all nodes lie to one side (typically right). If we perform an inorder traversal, we get a sorted list: [1, 2, 3, 4, 5]. We now take the middle element (in this case 3) and make it the root. The left half [1, 2] becomes the left subtree, and the right half [4, 5] becomes the right subtree.

We recursively repeat this process for each half. The result is a balanced BST where no side is significantly deeper than the other, and we maintain BST properties.

Case 2: Perfectly Balanced Input (e.g., [1, 2, 3, 4, 5, 6, 7])

If the input BST is already balanced, performing the balancing operation doesn't change its structure — though the algorithm still follows the same steps. The middle of [1, 2, 3, 4, 5, 6, 7] is 4, making it the root, and we recursively build the left subtree from [1, 2, 3] and the right from [5, 6, 7].

This case helps confirm that the algorithm doesn't break existing balance, and simply reuses a good structure.

Case 3: Single Node Tree (e.g., [5])

A BST with only one node is trivially balanced. The algorithm treats the only node as the root and does nothing more. It's an important edge case to verify your solution handles minimal input correctly.

Case 4: Empty Tree (e.g., [])

An empty tree is already balanced, as there's nothing to process. The algorithm should handle this gracefully by returning null or equivalent empty root, without errors or unnecessary operations.

Why This Approach Works

This approach ensures optimal balancing because each recursive step divides the tree into two halves, like a binary search. We always pick the middle value as the root, keeping both subtrees as equal in size as possible. This method guarantees a tree of minimal height and is widely used in balancing scenarios like AVL and Red-Black Trees.

Algorithm Steps❯

  1. Perform an inorder traversal of the BST to obtain a sorted array of node values.
  2. Recursively build a balanced BST from the sorted array:
  3. Find the middle element of the array and create a new node for it.
  4. Recursively repeat the process for the left subarray to build the left subtree and for the right subarray to build the right subtree.
  5. Return the newly constructed node as the root of the balanced BST.

Code

Python
Java
JavaScript
C
C++
Go
class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right


def inorder(root, arr):
    if root:
        inorder(root.left, arr)
        arr.append(root.val)
        inorder(root.right, arr)


def build_balanced_bst(sorted_arr):
    if not sorted_arr:
        return None
    mid = len(sorted_arr) // 2
    root = TreeNode(sorted_arr[mid])
    root.left = build_balanced_bst(sorted_arr[:mid])
    root.right = build_balanced_bst(sorted_arr[mid+1:])
    return root


def balance_bst(root):
    arr = []
    inorder(root, arr)
    return build_balanced_bst(arr)

# Example usage:
if __name__ == '__main__':
    # Construct an unbalanced BST (right-skewed)
    root = TreeNode(1, None, TreeNode(2, None, TreeNode(3, None, TreeNode(4))))
    balanced = balance_bst(root)
    arr = []
    inorder(balanced, arr)
    print(arr)