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

Construct a Binary Tree from a String with Bracket Representation

Problem Statement

You are given a string that represents a binary tree using bracket notation. Your task is to construct the binary tree from this string. The format of the string is as follows: - A node is represented by its value followed optionally by its left and right subtrees in parentheses. - For example, the string '4(2(3)(1))(6(5))' represents the following tree: 4 / \ 2 6 / \ / 3 1 5 Write a function to construct the binary tree from such a string. If the string is empty, return null.

Examples

Input Tree Level Order Output Description
"4(2(3)(1))(6(5))"
[[4], [2, 6], [3, 1, 5]] Standard binary tree with both left and right subtrees containing nested children
"1(2)(3)"
[[1], [2, 3]] Simple binary tree with one left and one right child
"1(2(3(4)))"
[[1], [2], [3], [4]] Tree skewed to the left only, with nested left children
"1()(2()(3))"
[[1], [2], [3]] Tree skewed to the right only using empty parentheses for null children
"7"
[[7]] Single-node tree with only root
[] Empty string input results in an empty tree

Visualization Player

Solution

Understanding the Problem

We are given a string that represents a binary tree using a bracketed structure. Each node is a number, and its children are enclosed in parentheses. For example, the string 4(2(3)(1))(6(5)) represents a binary tree where 4 is the root, 2 is the left child, and 6 is the right child. The left child 2 further has children 3 and 1. The task is to construct this binary tree from the string.

This problem is recursive in nature — every subtree is itself a valid string that can be parsed into a smaller binary tree. We need to handle parsing the string carefully, tracking open and closed parentheses to determine the subtree boundaries.

Step-by-Step Solution with Example

step 1: Identify the root value

We begin by extracting the integer value at the start of the string. This is the root node of our binary tree. For example, in 4(2(3)(1))(6(5)), we extract 4 as the root.

step 2: Locate the left and right subtrees

After the root, if the string continues with a '(', it signals the start of a subtree. We find the first matching pair of parentheses to extract the left subtree string. Then, we check if there's another set of parentheses for the right subtree.

In 4(2(3)(1))(6(5)), the first subtree is 2(3)(1), enclosed in the first pair of parentheses. The second subtree is 6(5), enclosed in the next pair.

step 3: Recursively construct left and right children

We recursively repeat the parsing process on each subtree string. The node 2 becomes the left child of 4. Similarly, within 2(3)(1), we identify 3 as the left child of 2 and 1 as the right child. For the right subtree 6(5), 6 becomes the right child of 4 and 5 becomes its left child.

step 4: Return the constructed tree node

Once the recursive calls return the constructed left and right subtrees, we attach them to the current node and return it. This process eventually constructs the full tree starting from the root.

Edge Cases

Case 1: Full Tree Structure (4(2(3)(1))(6(5)))

The input string has both left and right subtrees. We parse both sets of parentheses and recursively build each part.

Case 2: Only Left Subtree (1(2))

There is only one set of parentheses, indicating only a left child. The right child remains null.

Case 3: Only Right Subtree (1()(3))

An empty set of parentheses denotes a missing left child, and the next set represents the right child.

Case 4: Single Node Tree (1)

No parentheses means no children. The tree consists of a single root node with both children as null.

Case 5: Empty Input ("")

An empty string means there's no tree to build. We return null immediately — this is often a base case in recursive functions.

Finally

This problem is an excellent example of recursive string parsing and tree construction. The key is to understand how to locate matching parentheses and recursively build subtrees. By addressing edge cases like missing children and empty inputs, we ensure our solution is robust and handles all possible valid inputs gracefully.

Algorithm Steps

  1. Given a string representing a binary tree in bracket notation (e.g., 4(2(3)(1))(6(5))).
  2. Initialize an index pointer at the start of the string.
  3. Parse the numeric value (and sign, if negative) from the current index.
  4. Create a new tree node with the parsed value.
  5. If the next character is (, recursively construct the left subtree and skip the corresponding ).
  6. If another ( follows, recursively construct the right subtree and skip its closing ).
  7. Return the constructed node and update the index.
  8. Repeat the process until the entire string is parsed and the binary tree is built.

Code

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

def str2tree(s: str):
    if not s:
        return None
    def helper(i):
        # Parse number (handle negative numbers)
        sign = 1
        if s[i] == '-':
            sign = -1
            i += 1
        num = 0
        while i < len(s) and s[i].isdigit():
            num = num * 10 + int(s[i])
            i += 1
        node = TreeNode(sign * num)
        # Parse left subtree if '(' found
        if i < len(s) and s[i] == '(':
            i += 1  # skip '('
            node.left, i = helper(i)
            i += 1  # skip ')'
        # Parse right subtree if '(' found
        if i < len(s) and s[i] == '(':
            i += 1  # skip '('
            node.right, i = helper(i)
            i += 1  # skip ')'
        return node, i
    root, _ = helper(0)
    return root

# Example usage:
if __name__ == '__main__':
    s = "4(2(3)(1))(6(5))"
    root = str2tree(s)
    # A function to print the tree can be added for verification
    print(root.val)

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