Binary TreesBinary Trees36

  1. 1Preorder Traversal of a Binary Tree using Recursion
  2. 2Preorder Traversal of a Binary Tree using Iteration
  3. 3Postorder Traversal of a Binary Tree Using Recursion
  4. 4Postorder Traversal of a Binary Tree using Iteration
  5. 5Level Order Traversal of a Binary Tree using Recursion
  6. 6Level Order Traversal of a Binary Tree using Iteration
  7. 7Reverse Level Order Traversal of a Binary Tree using Iteration
  8. 8Reverse Level Order Traversal of a Binary Tree using Recursion
  9. 9Find Height of a Binary Tree
  10. 10Find Diameter of a Binary Tree
  11. 11Find Mirror of a Binary Tree - Todo
  12. 12Inorder Traversal of a Binary Tree using Recursion
  13. 13Inorder Traversal of a Binary Tree using Iteration
  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

Implement Atoi (String to Integer Conversion)



Problem Statement

You are given a string that may contain a number. Your task is to convert this string into a valid integer, just like the C/C++ function atoi() does.

However, the conversion should follow these rules:

If the string does not contain a valid number, return 0.

Examples

Input StringOutputDescription
"42"42Normal case with valid positive number
" -42"-42Leading spaces and negative sign handled
"4193 with words"4193Number ends before non-digit characters
"words and 987"0Invalid as it doesn't start with a number
"+123"123Valid positive number with explicit sign
"-91283472332"-2147483648Below 32-bit signed int range, clamped to INT_MIN
"91283472332"2147483647Above 32-bit signed int range, clamped to INT_MAX
""0Empty string returns 0
" "0Whitespace only returns 0
"-"0Only sign with no digits returns 0
"+0 123"0Only 0 is read before space, rest ignored

Solution

To convert a string into an integer (like the atoi function), we need to carefully interpret the string one character at a time. It's not just about parsing digits — we must also think about signs, spaces, invalid characters, and potential overflows.

Let’s break it down for a beginner:

1. Ignore leading whitespaces: The first thing we do is skip any space characters at the beginning of the string. These should not affect the result.

2. Check for optional '+' or '-' sign: After trimming spaces, if the first character is '+' or '-', we store this to determine the final sign of our number.

3. Convert following digits: Next, we move character by character and as long as we see digits (0-9), we build the number by multiplying the existing result by 10 and adding the new digit.

4. Stop on invalid character: If we hit a character that's not a digit (e.g., letter, punctuation, space), we stop immediately. Only the valid prefix of the string is considered.

5️⃣ Handle overflow: While building the number, if at any point it goes beyond the limits of a 32-bit signed integer, we stop and clamp the result to:

  • 2147483647 if the number is too large (positive overflow)
  • -2147483648 if the number is too small (negative overflow)

6️⃣ Empty or invalid string: If after removing spaces and optional sign we find nothing to convert, or the string starts with something invalid, we simply return 0.

Example thoughts:

  • "42" → clean numeric string, converts to 42.
  • " -42" → spaces skipped, '-' sign handled, digits converted → -42.
  • "words and 987" → starts with non-digit → invalid → return 0.
  • "91283472332" → valid number but too large → clamp to 2147483647.
  • "" or " " → empty after trimming → return 0.

This approach ensures correctness and protects against errors, making it suitable for real-world use cases like parsing user input.

Algorithm Steps

  1. Trim leading whitespaces from the input string
  2. Check if the next character is '+' or '-'. Store the sign
  3. Iterate through each character:
  4. → If it's a digit, convert and add to result
  5. → If not a digit, break the loop
  6. → Before each step, check if result will overflow/underflow
  7. Return result multiplied by sign

Code

Java
Python
JavaScript
C
C++
C#
Kotlin
Swift
Go
Php
public class AtoiConverter {
    public static int myAtoi(String s) {
        if (s == null || s.isEmpty()) return 0;
        s = s.trim(); // remove leading/trailing spaces
        if (s.isEmpty()) return 0;

        int sign = 1, i = 0, result = 0;
        int INT_MAX = Integer.MAX_VALUE;
        int INT_MIN = Integer.MIN_VALUE;

        if (s.charAt(i) == '+' || s.charAt(i) == '-') {
            sign = (s.charAt(i) == '-') ? -1 : 1;
            i++;
        }

        while (i < s.length() && Character.isDigit(s.charAt(i))) {
            int digit = s.charAt(i) - '0';

            if (result > (INT_MAX - digit) / 10) {
                return (sign == 1) ? INT_MAX : INT_MIN;
            }

            result = result * 10 + digit;
            i++;
        }

        return result * sign;
    }

    public static void main(String[] args) {
        String input = "   -42";
        System.out.println("Converted Integer: " + myAtoi(input));
    }
}

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(1)When the string is empty or contains only whitespaces, the function exits early.
Average CaseO(n)Where n is the number of characters to parse through until a non-digit or end is encountered.
Average CaseO(n)Involves scanning through the entire valid digit portion of the string for parsing.

Space Complexity

O(1)

Explanation: Only a fixed number of variables are used for computation, regardless of input size.



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