Binary TreesBinary Trees36
  1. 1Preorder Traversal of a Binary Tree using Recursion
  2. 2Preorder Traversal of a Binary Tree using Iteration
  3. 3Inorder Traversal of a Binary Tree using Recursion
  4. 4Inorder Traversal of a Binary Tree using Iteration
  5. 5Postorder Traversal of a Binary Tree Using Recursion
  6. 6Postorder Traversal of a Binary Tree using Iteration
  7. 7Level Order Traversal of a Binary Tree using Recursion
  8. 8Level Order Traversal of a Binary Tree using Iteration
  9. 9Reverse Level Order Traversal of a Binary Tree using Iteration
  10. 10Reverse Level Order Traversal of a Binary Tree using Recursion
  11. 11Find Height of a Binary Tree
  12. 12Find Diameter of a Binary Tree
  13. 13Find Mirror of a Binary Tree
  14. 14Left View of a Binary Tree
  15. 15Right View of a Binary Tree
  16. 16Top View of a Binary Tree
  17. 17Bottom View of a Binary Tree
  18. 18Zigzag Traversal of a Binary Tree
  19. 19Check if a Binary Tree is Balanced
  20. 20Diagonal Traversal of a Binary Tree
  21. 21Boundary Traversal of a Binary Tree
  22. 22Construct a Binary Tree from a String with Bracket Representation
  23. 23Convert a Binary Tree into a Doubly Linked List
  24. 24Convert a Binary Tree into a Sum Tree
  25. 25Find Minimum Swaps Required to Convert a Binary Tree into a BST
  26. 26Check if a Binary Tree is a Sum Tree
  27. 27Check if All Leaf Nodes are at the Same Level in a Binary Tree
  28. 28Lowest Common Ancestor (LCA) in a Binary Tree
  29. 29Solve the Tree Isomorphism Problem
  30. 30Check if a Binary Tree Contains Duplicate Subtrees of Size 2 or More
  31. 31Check if Two Binary Trees are Mirror Images
  32. 32Calculate the Sum of Nodes on the Longest Path from Root to Leaf in a Binary Tree
  33. 33Print All Paths in a Binary Tree with a Given Sum
  34. 34Find the Distance Between Two Nodes in a Binary Tree
  35. 35Find the kth Ancestor of a Node in a Binary Tree
  36. 36Find All Duplicate Subtrees in a Binary Tree
GraphsGraphs46
  1. 1Breadth-First Search in Graphs
  2. 2Depth-First Search in Graphs
  3. 3Number of Provinces in an Undirected Graph
  4. 4Connected Components in a Matrix
  5. 5Rotten Oranges Problem - BFS in Matrix
  6. 6Flood Fill Algorithm - Graph Based
  7. 7Detect Cycle in an Undirected Graph using DFS
  8. 8Detect Cycle in an Undirected Graph using BFS
  9. 9Distance of Nearest Cell Having 1 - Grid BFS
  10. 10Surrounded Regions in Matrix using Graph Traversal
  11. 11Number of Enclaves in Grid
  12. 12Word Ladder - Shortest Transformation using Graph
  13. 13Word Ladder II - All Shortest Transformation Sequences
  14. 14Number of Distinct Islands using DFS
  15. 15Check if a Graph is Bipartite using DFS
  16. 16Topological Sort Using DFS
  17. 17Topological Sort using Kahn's Algorithm
  18. 18Cycle Detection in Directed Graph using BFS
  19. 19Course Schedule - Task Ordering with Prerequisites
  20. 20Course Schedule 2 - Task Ordering Using Topological Sort
  21. 21Find Eventual Safe States in a Directed Graph
  22. 22Alien Dictionary Character Order
  23. 23Shortest Path in Undirected Graph with Unit Distance
  24. 24Shortest Path in DAG using Topological Sort
  25. 25Dijkstra's Algorithm Using Set - Shortest Path in Graph
  26. 26Dijkstra’s Algorithm Using Priority Queue
  27. 27Shortest Distance in a Binary Maze using BFS
  28. 28Path With Minimum Effort in Grid using Graphs
  29. 29Cheapest Flights Within K Stops - Graph Problem
  30. 30Number of Ways to Reach Destination in Shortest Time - Graph Problem
  31. 31Minimum Multiplications to Reach End - Graph BFS
  32. 32Bellman-Ford Algorithm for Shortest Paths
  33. 33Floyd Warshall Algorithm for All-Pairs Shortest Path
  34. 34Find the City With the Fewest Reachable Neighbours
  35. 35Minimum Spanning Tree in Graphs
  36. 36Prim's Algorithm for Minimum Spanning Tree
  37. 37Disjoint Set (Union-Find) with Union by Rank and Path Compression
  38. 38Kruskal's Algorithm - Minimum Spanning Tree
  39. 39Minimum Operations to Make Network Connected
  40. 40Most Stones Removed with Same Row or Column
  41. 41Accounts Merge Problem using Disjoint Set Union
  42. 42Number of Islands II - Online Queries using DSU
  43. 43Making a Large Island Using DSU
  44. 44Bridges in Graph using Tarjan's Algorithm
  45. 45Articulation Points in Graphs
  46. 46Strongly Connected Components using Kosaraju's Algorithm

Balance a Binary Search Tree - Algorithm, Visualization, Examples

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)