Learn and practice the Tree Depth-First Search, Tree Breadth-First Search, and Backtracking patterns to solve coding interview problems.

Grokking the Blind 75 Problems in Go

In this module, we’ll learn to work with the tree data structure and explore the use of the backtracking algorithm. Trees play a pivotal role in organizing hierarchical data, optimizing searches, and solving complex problems. We’ll learn to choose the appropriate tree for a given problem and apply the more appropriate of the two major traversal techniques, breadth-first or depth-first, to solve it. We use backtracking when seeking optimal solutions in a scenario where no possibility may be left unexplored. We’ll gain experience with using tree depth-first search to implement backtracking when the solution space is represented as a tree, and we’ll need to exhaustively explore all possible solutions. We’ll also learn to implement backtracking to solve problems where the solution space is represented using other data structures.

Exploring the Solution Space

# Statement

Given the root of a binary tree, display the values of its nodes while performing a level order traversal. Return the node values for all levels in a string separated by the character `:`. If the tree is empty, i.e., the number of nodes is $0$, then return "None" as the output.

**Constraints:**

- The number of nodes in the tree is in the range $[0, 500]$.

- $-10^3 \leq$ `Node.data` $\leq 10^3$

# Examples

# Understand the problem
Let’s take a moment to make sure you’ve correctly understood the problem. The quiz below helps you check if you’re solving the correct problem:



By using a stack data structure to keep track of visited nodes

By using a queue data structure to maintain the order of node exploration

By recursively traversing the left and right subtrees in a depth-first manner

 By randomizing the order of node exploration for better coverage

How does BFS ensure that it visits all nodes in a binary tree while traversing level by level?

How does BFS ensure that it visits all nodes in a binary tree while traversing level by level?

What is the output if the following tree is given as input?

```text
         ___ 50 _ 
	    |        |
	    -50 _    200 
	         |
	      __ 32 
	     |
	  __ 12 
	 |
	 -1 

What is the output if the following tree is given as input?
<pre><code class="language-text"> ___ 50 _ 
	 | |
	 -50 _ 200 
	 |
	 __ 32 
	 |
	 __ 12 
	 |
	 -1 </code></pre>

What is the output if the following tree is given as input?

```text
	         __ 50 _ 
	        |       |
	     __ 20      60 _ 
	    |               |
	  _ 10              70 _ 
	 |                      |
	 5                      80 _ 
	                            |
	                            90 

What is the output if the following tree is given as input?
<pre><code class="language-text">	 __ 50 _ 
	 | |
	 __ 20 60 _ 
	 | |
	 _ 10 70 _ 
	 | |
	 5 80 _ 
	 |
	 90 </code></pre>

# Figure it out!
We have a game for you to play. Rearrange the logical building blocks to develop a clearer understanding of how to solve this problem.

# Try it yourself
Implement your solution in `main.go` in the following coding playground. You'll need the provided supporting code to implement your solution.

> **Note:** The binary tree node's class has members `left` and `right` to store references to other nodes, along with the member `data` to hold the node's value.

package main
import (. "golang-test-code/ds_v1/BinaryTree") // TreeNode[T]

import (
	"strconv"
)

// Queue is used by binary tree

// Node type struct
type Node struct {
	value *TreeNode[int]
	next  *Node
}

// Queue type struct
type Queue struct {
	head *Node
	tail *Node
	size int
}

// Size returns the size of the queue
func (q *Queue) Size() int {
	return q.size
}

// IsEmpty checks whether the queue is empty or not
func (q *Queue) IsEmpty() bool {
	return q.size == 0
}

// Peek returns the top element of the queue
func (q *Queue) Peek() *TreeNode[int] {
	if q.IsEmpty() {
		return nil
	}
	return q.head.value
}

// Enqueue push an element into the queue
func (q *Queue) Enqueue(data *TreeNode[int]) {
	temp := &Node{data, nil}
	if q.head == nil {
		q.head = temp
		q.tail = temp
	} else {
		q.tail.next = temp
		q.tail = temp
	}
	q.size++
}

// Dequeue remove an element from the queue
func (q *Queue) Dequeue() *TreeNode[int] {
	if q.IsEmpty() {
		return nil
	}
	value := q.head.value
	q.head = q.head.next
	q.size--
	return value
}

// String function converts queue elements to string
func (q *Queue) String() string {
	output := "["
	if q.head != nil{
		temp := q.head
		for temp != nil {
			output += strconv.Itoa(temp.value.Data) + ", "
			temp = temp.next
		}
		output = output[0:len(output)-2]
	}
	output += "]"
	return  output
}

package main

type BinaryTreeNode struct {
	data  int
	left  *BinaryTreeNode
	right *BinaryTreeNode

	// below data members used only for some of the problems
    next *BinaryTreeNode
	parent *BinaryTreeNode
	count int
}

// NewBinaryTreeNode initializes and returns the BinaryTreeNode 
// type object
func NewBinaryTreeNode(data int) *BinaryTreeNode {
	t := new(BinaryTreeNode)
	t.data = data
	t.left = nil
	t.right = nil
	t.next = nil
	t.parent = nil
	t.count = 0
	return t
}

# Quick Fix\nYou are Edward, a Senior Software Engineer who specializes in coding interview patterns, data structures and algorithms, and is great at debugging issues in code.\n\n- Edward, please keep in mind that you do not have to print the data of the placeholders. Also, do not mention anything about yourself.\n\nEdward, you are given the following data in the placeholders:\n- Function name: `levelOrderTraversal`\n- Candidate's submitted solution code: USER_CODE\n- Error: ERROR_DESCRIPTION\n- Failed test case: FAILED_TEST_CASE\n- Coding language: `Go`\n\n- Edward, your task is to understand the type of error that is occurring in the coding language provided.\n- Once you've understood the error as per the error’s description, and why the test case failed, you will suggest how the candidate can improve their solution code to fix the error.\n- Edward, you will not give the corrected code, just a hint on what the error tells, and how to fix it.\n- Edward, please limit your response strictly to 100 words. No more words than the limit.

# Review Code\nEdward, your current task is to review the `levelOrderTraversal` function(s) and the other function(s) invoked in it/them.\n\n-Take your time and review the candidate's solution code.\n- Do not review any extra/helper files or functions not invoked in the solution function `levelOrderTraversal`.\n- Do not review the code of data structure objects instantiated like a binary tree, linked list, trie, union find, etc., in the solution functions. Just review the solution/algorithm function of the problem. \n- Also, if the solution code doesn’t use the coding pattern stated above, then please encourage the candidate to reattempt the problem using that pattern corectly.

# Review time complexity\n- Please provide a brief time complexity analysis of the function(s) `levelOrderTraversal`. \n- While calculating time complexity, include the time complexity of built-in Go functions invoked by the code. \n- The answer should be to the point in approximately 60 words.

# Review space complexity\n- Please provide a brief space complexity analysis of the function(s) `levelOrderTraversal`. \n- While calculating space complexity, include the space complexity of built-in Go functions invoked by the code. \n- Do not include the space used for the input parameters and the space used for just returning an output (not used as a scratch space). \n- The answer should be to the point in approximately 60 words.

def evaluate_results(self) -> object:
 """
 self.__inputs is the list of inputs that are passed as input to function
 self.__actual_output is a list of outputs that is produced by the learner's code
 return : self.__edu__result object
 """
 # multiple outputs will be added at each index respectively for a single test_case

 # from ds_v1.BinaryTree.BinaryTree import TreeNode
 class TreeNode:
 def __init__(self, data):
 self.data = data
 self.left = None
 self.right = None
 from collections import deque

 def level_order_traversal(root):
 result = ""
 # Print 'None' if the root is empty
 if not root:
 result = "None"
 return result
 else:
 # Initializing the current queue
 current_queue = deque()

 # Initializing the dummy node
 dummy_node = TreeNode(0)

 current_queue.append(root)
 current_queue.append(dummy_node)

 # Printing nodes in level-order until the current queue remains
 # empty
 while current_queue:
 # Dequeuing and printing the first element of queue
 temp = current_queue.popleft()
 result += str(temp.data)

 # Adding dequeued node's children to the next queue
 if temp.left:
 current_queue.append(temp.left)

 if temp.right:
 current_queue.append(temp.right)

 # When the dummyNode comes next in line, we print a new line and dequeue
 # it from the queue
 if current_queue[0] == dummy_node:
 current_queue.popleft()

 # If the queue is still not empty we add back the dummy node
 if current_queue:
 result += " : "
 current_queue.append(dummy_node)
 else:
 result += ", "
 return result
 

 def create_binary_tree(level_order):
 if not level_order:
 return None
 
 root = TreeNode(level_order[0])
 q = deque([root])
 i = 1
 while i < len(level_order) and q:
 curr = q.popleft()
 
 if level_order[i] != None:
 curr.left = TreeNode(level_order[i])
 q.append(curr.left)
 i += 1
 
 if i < len(level_order) and level_order[i] != None:
 curr.right = TreeNode(level_order[i])
 q.append(curr.right)
 i += 1
 
 return root

 expected_outputs = [] 
 # store the 1st expected output that should be displayed to the user 
 inp=self.__inputs[0]
 root=create_binary_tree(inp)
 out=level_order_traversal(root)





 expected_outputs.append(out)

 if json.dumps(out) == json.dumps(self.__actual_outputs[0]):
 self.__edu_result.store_result(expected_outputs, TestCaseResult.PASS)
 else:
 self.__edu_result.store_result(expected_outputs, TestCaseResult.FAIL)

 return self.__edu_result

// Definition of a binary tree node 
// type TreeNode[T any] struct {
// 	Data  T
// 	Left  *TreeNode[T]
// 	Right *TreeNode[T]
// }

package main
import (. "golang-test-code/ds_v1/BinaryTree") // TreeNode[T]

func levelOrderTraversal(root *TreeNode[int]) string{
    
    // Replace this placeholder return statement with your code
    return ""
}

level_order_traversal.yaml

testcases:
  - name: "testcase 1"
    inputs:
    - 1 : [100, 50, 200, 25, 75, 300, 10, 350, 15]
    output:
    - 1 : "100 : 50, 200 : 25, 75, 300, 10 : 350, 15"

  - name: "testcase 2"
    inputs:
    - 1 : [25, 50, 75, 100, 200, 350]
    output:
    - 1 : "25 : 50, 75 : 100, 200, 350"

  - name: "testcase 3"
    inputs:
    - 1 : []
    output:
    - 1 : "None"

  - name: "testcase 4"
    inputs:
    - 1 : [100]
    output:
    - 1 : "100"

  - name: "testcase 5"
    inputs:
    - 1 : [350, 200, null, 75, 50, 25]
    output:
    - 1 : "350 : 200 : 75, 50 : 25"

def sum(input_1, input_2):
    return input_1+input_2

def validate_inputs(self) -> None:
 """
 self.__inputs is the list of inputs that are passed as input to function
 store the result using the self.store_result(error_msg, status)
 return : void
 """
 # inputs can be accessed using self.__inputs starting from 0 to len(self.__inputs) - 1


 class TreeNode:
 def __init__(self, data):
 self.data = data
 self.left = None
 self.right = None
 from collections import deque

 def level_order_traversal(root):
 if not root:
 return []
 
 result = []
 q = deque([root])
 while q:
 curr = q.popleft()
 
 if curr:
 result.append(curr.data)
 q.append(curr.left)
 q.append(curr.right)
 else:
 result.append(None)

 if all(val == None for val in q):
 break 
 
 return result
 
 def is_valid_level_order(level_order):
 if not level_order:
 return True # An empty list is a valid level-order traversal

 n = len(level_order)
 i = 0
 root = TreeNode(level_order[i])
 q = deque([root])
 i = 1

 while i < n and q:
 curr = q.popleft()

 if level_order[i] is not None:
 curr.left = TreeNode(level_order[i])
 q.append(curr.left)
 i += 1

 if i < n and level_order[i] is not None:
 curr.right = TreeNode(level_order[i])
 q.append(curr.right)
 i += 1
 
 # Remove None values after the last integer
 while level_order and level_order[-1] is None:
 level_order.pop()
 
 return level_order_traversal(root) == level_order


 tree=self.__inputs[0]
 if not is_valid_level_order(tree):
 self.store_result ("Not a valid value of type TreeNode", ValidationFunction.FAIL)
 elif not all(-1000 <= num <= 1000 for num in tree if num is not None):
 self.store_result("The tree node should consist of values ranging from -1000 to 1000.", ValidationFunction.FAIL)

 elif (len(tree) > 0 and tree[0] is None):
 self.store_result ("Not a valid value of type TreeNode", ValidationFunction.FAIL)


 elif not (0 <= sum(1 for value in tree if value is not None) <= 500):
 self.store_result("The tree should have number of nodes in the range 0 to 500.", ValidationFunction.FAIL)
 
 else:
 self.store_result("", ValidationFunction.PASS)

Try to solve the Binary Tree Level Order Traversal problem.

Tree Breadth-first Search

Tree Depth-first Search

Backtracking

Conclusion

Level Order Traversal of Binary Tree

Statement

Examples

Understand the problem

Figure it out!

Try it yourself