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

Speedrun the Meta Coding Interview: Top Problems in JavaScript

Trees play a pivotal role in organizing hierarchical data, optimizing searches, and solving complex problems. In this module, 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.

Trees

# Statement
Given the `root` of a binary tree, the task is to flatten the tree into a linked list using the same `TreeNode` class. The left child pointer of each node in the linked list should always be NULL, and the right child pointer should point to the next node in the linked list. The nodes in the linked list should be in the same order as that of the preorder traversal of the given binary tree.

**Constraints:** 

* $-100 \leq$ `Node.data` $\leq 100$.
* The tree contains nodes in the range $[1, 500]$.

# 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:


3 → 2 → 17 → 1 → 4 → 19 → 5



3 → 2 → 1 → 4 → 17 → 19 → 5



3 → 2 → 1 → 4 → 19 → 17 → 5



17 → 19 → 5 → 3 → 2 → 1 → 4


Choose the correct flattened linked list for the given binary tree:
```text
    3
   / \
  2   17
 / \  / \
1  4 19  5
```

Choose the correct flattened linked list for the given binary tree:
<pre><code class="language-text"> 3
 / \
 2 17
 / \ / \
1 4 19 5
</code></pre>

Choose the correct flattened linked list for the given binary tree:
```text
    -1
    / \
  -2   -5
  / \    \
-3  -4   -6
```

Choose the correct flattened linked list for the given binary tree:
<pre><code class="language-text"> -1
 / \
 -2 -5
 / \ \
-3 -4 -6
</code></pre>

Flatten Binary Tree to Linked List

Flatten Binary Tree to Linked List

# 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.js` in the following coding playground.

from binary_tree_node import *

class BinaryTree:
 def __init__(self, *args):
 if len(args) < 1:
 self.root = None
 elif isinstance(args[0], int):
 self.root = BinaryTreeNode(args[0])
 else:
 self.root = None
 for x in args[0]:
 self.insert(x)

 # for BST insertion
 def insert(self, node_data):
 new_node = BinaryTreeNode(node_data)
 if not self.root:
 self.root = new_node
 else:
 parent = None
 temp_pointer = self.root
 while temp_pointer:
 parent = temp_pointer
 if node_data <= temp_pointer.data:
 temp_pointer = temp_pointer.left
 else:
 temp_pointer = temp_pointer.right
 if node_data <= parent.data:
 parent.left = new_node
 else:
 parent.right = new_node

 def find_in_bst_rec(self, node, node_data):
 if not node:
 return None
 if node.data == node_data:
 return node
 elif node.data > node_data:
 return self.find_in_bst_rec(node.left, node_data)
 else:
 return self.find_in_bst_rec(node.right, node_data)

 def find_in_bst(self, node_data):
 return self.find_in_bst_rec(self.root, node_data)

 def get_sub_tree_node_count(self, node):
 if not node:
 return 0
 else:
 return 1 + self.get_sub_tree_node_count(node.left) + self.get_sub_tree_node_count(node.right)

 def get_tree_deep_copy_rec(self, node):
 if node:
 new_node = BinaryTreeNode(node.data)
 new_node.left = self.get_tree_deep_copy_rec(node.left)
 new_node.right = self.get_tree_deep_copy_rec(node.right)
 return new_node
 else:
 return None

 def get_tree_deep_copy(self):
 if not self.root:
 return None
 else:
 tree_copy = BinaryTree()
 tree_copy.root = self.get_tree_deep_copy_rec(self.root)
 return tree_copy

# Template for binary tree node

class BinaryTreeNode:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None

        # below data members used only for some of the problems
        self.next = None
        self.parent = None
        self.count = 0

# 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: `flattenTree`\n- Candidate's submitted solution code: USER_CODE\n- Error: ERROR_DESCRIPTION\n- Failed test case: FAILED_TEST_CASE\n- Coding language: `JavaScript`\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 `flattenTree` 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 `flattenTree`.\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) `flattenTree`. \n- While calculating time complexity, include the time complexity of built-in JavaScript 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) `flattenTree`. \n- While calculating space complexity, include the space complexity of built-in JavaScript 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
 """
 
 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 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
 
 def flatten_tree(root):
 if not root:
 return
 
 current = root
 while current:

 if current.left:
 last = current.left

 while last.right:
 last = last.right
 
 last.right = current.right
 current.right = current.left
 current.left = None

 current = current.right
 return root

 # multiple outputs will be added at each index respectively for a single test_case
 expected_outputs = [] 
 inp = self.__inputs[0]
 inp_root = create_binary_tree(inp)
 expected1 = flatten_tree(inp_root)
 expected = level_order_traversal(expected1)
 expected_outputs.append(expected)
 actual1 = self.__actual_outputs[0]
 
 if json.dumps(expected) == json.dumps(actual1):
 self.__edu_result.store_result(expected_outputs, TestCaseResult.PASS)
 else:
 self.__edu_result.store_result(expected_outputs, TestCaseResult.FAIL)

 return self.__edu_result

class TreeNode:
 def __init__(self, data):
 self.data = data
 self.left = None
 self.right = None

import {TreeNode} from './ds_v1/BinaryTree.js';

// Definition of a binary tree node
// class TreeNode {
//   constructor(data) {
//     this.data = data;
//     this.left = null;
//     this.right = null;
//   }
// }

export function flattenTree(root) {

    // Replace this placeholder return statement with your code
    return root;
}


flatten_tree.yaml

testcases:
  - name: "testcase 1"
    inputs:
    - 1 : [3, 2, 17, 1, 4, 19, 5]
    output:
    - 1 : [3, null, 2, null, 1, null, 4, null, 17, null, 19, null, 5]

  - name: "testcase 2"
    inputs:
    - 1 : [1, 2, 3, 4, 5, 6, 7]
    output:
    - 1 : [1, null, 2, null, 4, null, 5, null, 3, null, 6, null, 7]

  - name: "testcase 3"
    inputs:
    - 1 : [1, 2, 3, 4, 5, 6, 9, 7, 8]
    output:
    - 1 : [1, null, 2, null, 4, null, 7, null, 8, null, 5, null, 3, null, 6, null, 9]

  - name: "testcase 4"
    inputs:
    - 1 : [5, 2, 1, 6, 10, 11, 44]
    output:
    - 1 : [5, null, 2, null, 6, null, 10, null, 1, null, 11, null, 44]

  - name: "testcase 5"
    inputs:
    - 1 : [1, 2, 5, 3, 4, 6]
    output:
    - 1 : [1, null, 2, null, 3, null, 4, null, 5, null, 6]

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
 """
 
 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

 # inputs can be accessed using self.__inputs starting from 0 to len(self.__inputs) - 1
 input_list = self.__inputs[0]
 # inputs can be accessed using self.__inputs starting from 0 to len(self.__inputs) - 1
 if not all(-100 <= num <= 100 for num in input_list if num is not None):
 self.store_result ("'Node' must consist of values from -100 to 100 only", ValidationFunction.FAIL)
 elif (len(input_list) != 0 and input_list[0] is None):
 self.store_result ("Not a valid value of type TreeNode", ValidationFunction.FAIL)
 elif not (1 <= sum(1 for value in input_list if value is not None) <= 500):
 self.store_result ("Total nodes in the tree are expected to be between 1 and 500", ValidationFunction.FAIL)
 elif not is_valid_level_order(input_list):
 self.store_result ("Not a valid value of type TreeNode", ValidationFunction.FAIL)

 else:
 self.store_result("", ValidationFunction.PASS)

class TreeNode:
 def __init__(self, data):
 self.data = data
 self.left = None
 self.right = None

Try to solve the Flatten Binary Tree to Linked List problem.

Tree Depth-first Search

Tree Breadth-first Search

Conclusion

Flatten Binary Tree to Linked List

Statement

Examples

Understand the problem

Figure it out!

Try it yourself