Summary

This context discusses the implementation of Depth-First Search (DFS) and Breadth-First Search (BFS) algorithms in solving a LeetCode problem, Binary Tree Paths, using Golang, and the interconnections between DFS, BFS, and pre-order, post-order, and level-order traversals.

Abstract

The context begins by introducing the problem of finding all root-to-leaf paths in a binary tree, as presented in LeetCode problem 257. It then proceeds to provide solutions using DFS and BFS algorithms, demonstrating the use of recursion and stack for DFS, and queue for BFS. The solutions are further categorized into post-order and pre-order traversals for DFS, and level-order traversal for BFS. The context also discusses the time and space complexities of each solution and concludes by summarizing the key features of DFS and BFS algorithms.

Bullet points

The problem discussed is LeetCode problem 257: Binary Tree Paths.
The solutions provided use Depth-First Search (DFS) and Breadth-First Search (BFS) algorithms.
DFS solutions are implemented using recursion and stack, with post-order and pre-order traversals.
BFS solution is implemented using a queue and level-order traversal.
The time complexity for all solutions is O(n * h), where n is the number of nodes and h is the height of the tree.
The space complexity for DFS solutions is O(n * h), while for BFS it is O(n).
The context concludes by summarizing the key features of DFS and BFS algorithms.

LeetCode 257(Golang): Binary Tree Paths(Easy): Recursion vs BFS vs DFS

You understand concepts like DFS(Depth First Search), BFS(Breadth-First Search), etc., but do you know how they relate to pre-order, post-order, and level-order traversals? Let’s delve deeper into their interconnections through a simple LeetCode problem.

1. Problem 2. Solution && Analysis 4. Conclusion 5. Refefence

So let’s start :)

Problem

problem link

Given the root of a binary tree, return all root-to-leaf paths in any order.

A leaf is a node with no children.

Example 1:

Input: root = [1,2,3,null,5]
Output: ["1->2->5","1->3"]

Example 2:

Input: root = [1]
Output: ["1"]

Constraints:

The number of nodes in the tree is in the range [1, 100].
-100 <= Node.val <= 100

Solution && Analysis

DFS(Depth First Search)

Recursion && bottom-up && post-order traversal

From the first example, you can draw the following diagram:

> Why DFS? > Recursion is the ”stack“, it involves a stack pushing process, conforming to the stack’s Last-In-First-Out (LIFO) principle. > > Why post-order traversal and bottom-up？ > because we are the last to deal with the root node.

Time Complexity: O(n * h) Each node in the tree is visited once (n is the number of nodes). For each node, a string concatenation is performed, which can take up to h operations in the worst case, where h is the height of the tree.

Space Complexity: O(n * h) The space complexity is mainly due to the recursive call stack and the storage of path strings. In the worst case, the call stack can go as deep as the height of the tree, h, and for each call, a path string of length up to h is created. This leads to O(n * h) space usage, considering the worst-case scenario for a skewed tree.

2. Stack && up-bottom && preorder traversal

diagram:

> Why up-bottom && preorder traversal? > Handle the root node first.

Time Complexity: O(n * h) The time complexity is O(n * h) where n is the number of nodes in the tree, and h is the height of the tree. This is because each node is visited once, and for each node, path strings are concatenated and copied, which can take up to h operations in the worst case. Space Complexity: O(n) The space complexity is primarily affected by the storage of the stack and the path strings within each `StackNode`. 1. The stack itself, in the worst case, can have as many as `n` `StackNode` entries (especially in a skewed tree where the traversal resembles a linear list). 2. Each `StackNode` contains a path string, which can grow up to the height of the tree, `h`, in length.

BFS(Breadth-First Search)

level order traversal

diagram:

> Why BFS？ > Use of Queue: BFS is characterized by the use of a queue data structure to keep track of the nodes to be visited.

Time Complexity: O(n * h) Each node in the tree is visited once, where n is the number of nodes. The major cost comes from constructing the path strings for each node. Since paths are being concatenated at each level, the time to construct paths can be up to the height of the tree (h) for each node, leading to a worst-case time complexity of O(n * h). Space Complexity: O(n * h) 1. The space complexity is primarily due to the queue and the storage of path strings within each `nodePath` struct. 2. The queue may contain multiple `nodePath` structs, each with a path string. In the worst case, the length of these path strings can grow up to the height of the tree, h, especially for nodes deeper in the tree.

Conclusion

1. Depth-First Search (DFS) can generally be implemented in two ways: recursively or iteratively. It primarily leverages the characteristics of a stack, and can be combined with the concepts of pre-order and post-order traversals for top-down and bottom-up approaches, respectively. 2. Breadth-First Search (BFS) typically requires the integration of a queue for implementation and can be aligned with the ideas of level-order traversal 3. Commonly, when tackling a DFS or BFS problem, these features can be your starting points, enabling a systematic approach towards problem-solving. 4. In Go, the string type is immutable. Utilizing strings.Builder can be more efficient. If there’s interest in this area, I am also planning to write an article on this topic. Your comments and engagement are welcome.

Outstanding Question: Are there solutions as follows?

1. Recursion && up-bottom && Preorder traversal && DFS 2. Stack && bottom-up && Preorder traversal && DFS

Feel free to reply in the comments section. :)

Reference

LeetCode Difference between BFS and DFS — GeeksforGeeks