Maximum Depth of N-ary Tree

To find the maximum depth of an n-ary tree, we can use a recursive function that explores all the children of a node.

Code

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11

class Solution:
    def maxDepth(self, root: 'Node') -> int:
        if root is None:
            return 0

        max_depth = 0
        for child in root.children:
            if child:
                max_depth = max(max_depth, self.maxDepth(child))

        return max_depth + 1

Explanation

  1. Base Case: If the root is None, the depth is 0.
  2. Recursive Case: For each child of the root, call the maxDepth function recursively.
  3. Calculate Maximum Depth: Keep track of the maximum depth returned from the children, and add 1 for the current node.
  4. Return Result: Return the calculated maximum depth.

The time complexity of this approach is O(N), where N is the total number of nodes in the tree, as we visit each node once. The space complexity is O(H), where H is the height of the tree, due to the recursive call stack.

Define the Terms

N-ary Tree: A node can have many children.

Define the Interface Input: root reference of the n-ary tree Output: integer

Analyze the Input and Output Edge case : 0 nodes, return 0

Identify the Invariants

Identify the constraints

  • The depth of the n-ary tree is less than or equal to 1000.
  • The total number of nodes is between [0, 10^4].

Search the Problem Space

Classify the Problem

Analyze the Examples

Solve the Problem by Hand

Describe the Pattern

Distill the Observations to Create Insights

Brute Force Approach

Analyze Time and Space Complexity

Time: O( ) Space: O( )

Outline the Solution

Keep track of two things:

  • depth (current depth)
  • maximum depth (the final answer) Start from the root. current_depth = 1 Get all the children of the root. current_depth += 1 (the root must have at least one child) Iterate over all the children Which child will give

Recursive Approach

  1. The base case is when the root is null, return 0

  2. Loop through the children (root.children)

    • Update the current_depth by one more than its previous
    • Update the max_depth

    We can either have the max_depth as global variable or as a parameter

    Before the loop begins, ans = max(current_depth, max_depth) max_depth can be global

    If I am a leaf node, check the max and if my depth is higher update the depth otherwise, return from the recursive call.

    Unit of Work

    • Updating the max depth we have seen so far.
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14

class Solution:
    def dfs(self,node,depth):
        if(node is None):
            return 0

        self.ans = max(self.ans,depth)

        for child in node.children:
            self.dfs(child,depth+1)

    def maxDepth(self, root: 'Node') -> int:
        self.ans = 0
        self.dfs(root,1)
        return self.ans

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16

int calcDepth;
    int maxDepth(Node* root) {
        dfs(root, 0);
        return calcDepth;
    }
    private:
    void dfs(Node *root, int depth)
    {
        if(root == NULL) return ;

        for( auto &node:root->children)
            dfs(node, depth + 1);

        calcDepth = max(depth+1, calcDepth);
        return ;
    }

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27

# Definition for a Node.
# class Node
#     attr_accessor :val, :children
#     def initialize(val)
#         @val = val
#         @children = []
#     end
# end
def dfs(root, depth)
   if root.nil?
       return 
   end

   @max_depth = [depth, @max_depth].max

   root.children.each do |node|
      dfs(node, depth+1)   
   end
end

# @param {Node} root
# @return {int}
def maxDepth(root)
  @max_depth = 0
  dfs(root, 1)
  @max_depth
end

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15

class Solution:
    def maxDepth(self, root: 'Node') -> int:
        if not root:
            return 0

        nodes = deque()
        nodes.append((root, 1))
        maxx = 0
        while nodes:
            cur, val = nodes.popleft()
            maxx = val
            if cur.children:
                for child in cur.children:
                    nodes.append((child, val+1))
        return maxx

Problem Classification

This problem is related to Trees. It deals with the traversal and analysis of a n-ary tree, which is a form of hierarchical data structure where each node can have n number of children.

‘What’ components of this problem are:

  1. N-ary Tree: A n-ary tree is a tree data structure where each node can have no more than n children.

  2. Maximum Depth: This is the length of the longest path from the root node down to the furthest leaf node.

  3. Nary-Tree Input Serialization: The problem states that the input is represented as a level order traversal of the tree, where each group of children is separated by the null value.

The problem can further be classified into the Tree Traversal sub-domain within Data Structures. It requires understanding and implementing tree traversal algorithms to identify and return the maximum depth of the tree. It implies using a depth-first search (DFS) or breadth-first search (BFS) traversal approach to explore all paths from the root to the leaf nodes, tracking the length of each path and returning the maximum length.

  • Domain: Trees, Tree Traversal
  • Main components: N-ary tree, Maximum depth, Tree traversal (level order traversal represented by Nary-Tree Input Serialization)

Language Agnostic Coding Drills

  1. Dissect the Code

Here are the distinct concepts and sub-concepts contained in the code:

a. Function Definitions: The syntax and structure of defining a function in a programming language.

b. Base Case Handling: Checking the root of the tree to see if it’s null, and returning an appropriate value (0 in this case) if it is.

c. Usage of Data Structures: The implementation uses a queue to handle the traversal of the tree nodes. Understanding and implementing queues are fundamental here.

d. Looping: The use of a while loop to traverse through all elements until the condition is no longer met.

e. Unpacking: The concept of unpacking a data structure into multiple variables in a single line.

f. Conditional Checking: Checking if a node has children, using an if statement.

g. Traversing Children Nodes: The ability to traverse through the children of a node.

h. Updating Data Structures: Adding elements to the end of a queue.

i. Keeping Track of Depth: The concept of keeping track of the maximum depth by storing the value of the maximum depth encountered during the traversal.

  1. Ordered List of Concepts

    a. Function Definitions (Easy): Fundamental to all programming tasks.

    b. Base Case Handling (Easy): An important concept in recursion and when dealing with trees, but fairly straightforward.

    c. Usage of Data Structures - Queues (Intermediate): Understanding queues is a fundamental part of data structures, but applying them effectively can be challenging.

    d. Looping (Easy): Looping is a fundamental concept in programming.

    e. Unpacking (Easy): Unpacking is a helpful tool in many languages that makes code cleaner and easier to read.

    f. Conditional Checking (Easy): Fundamental concept used in all forms of programming.

    g. Traversing Children Nodes (Intermediate): This requires understanding of the tree data structure and how to iterate over it.

    h. Updating Data Structures - Adding to Queue (Easy): Basic understanding of how to interact with data structures.

    i. Keeping Track of Depth (Intermediate): While not complex, this concept does require an understanding of the problem domain and some foresight to implement effectively.

  2. Problem-solving Approach

The solution uses a breadth-first search (BFS) approach, which is a common method for tree traversal. Each node of the tree is paired with its depth level in a tuple and added to a queue. The BFS proceeds by dequeuing a node and its depth, then enqueuing all its children paired with the updated depth (parent’s depth + 1).

In a BFS traversal, the queue always contains nodes at the current level before any nodes at the next level. So, the last node dequeued is guaranteed to be the one at the maximum depth, i.e., the deepest node encountered during the traversal. By keeping track of this node’s depth, we effectively track the maximum depth of the tree.

This approach ensures that each node is visited exactly once and that its depth compared to the root is properly computed. This leads us to the final solution: the maximum depth of the n-ary tree.

Targeted Drills in Python

  1. Python-based Coding Drills

Let’s write Python-based coding drills to encapsulate each identified concept.

a. Function Definitions

1
2

def example_function():
    print("This is an example function!")

b. Base Case Handling

1
2
3
4
5

def check_null(value):
    if value is None:
        return True
    else:
        return False

c. Usage of Data Structures - Queues

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11

from collections import deque

# Initialize a queue
queue = deque()

# Append elements
queue.append("Python")
queue.append("Java")

# Pop an element
queue.popleft() # Output: "Python"

d. Looping

1
2

for i in range(5):
    print(i)

e. Unpacking

1
2
3
4

pair = ('apple', 'banana')
fruit1, fruit2 = pair
print(fruit1)  # Output: "apple"
print(fruit2)  # Output: "banana"

f. Conditional Checking

1
2
3

value = 10
if value > 5:
    print("The value is greater than 5")

g. Traversing Children Nodes

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11

class Node:
    def __init__(self, value, children=[]):
        self.value = value
        self.children = children

# Create node with children
root = Node(1, [Node(2), Node(3)])

# Access children
for child in root.children:
    print(child.value)  # Output: 2, 3

h. Updating Data Structures - Adding to Queue

1
2
3
4

from collections import deque
queue = deque()

queue.append('Python')  # Add element to the queue

i. Keeping Track of Depth

1
2
3
4

depth = 0
for i in range(10):
    # Increase depth
    depth += 1
  1. Problem-specific Drills

For this particular problem, we need to understand how to traverse an n-ary tree using Breadth-First Search (BFS).

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11

from collections import deque

def bfs(root):
    if root is None:
        return

    queue = deque([root])
    while queue:
        node = queue.popleft()
        print(node.value)
        queue.extend(node.children)
  1. Integration of Drills

To solve the problem, we integrate the above drills in the following way:

  1. We start with a function definition maxDepth that takes the root of the tree as an argument.
  2. We handle the base case by checking if the root is None. If so, we return 0.
  3. We initialize a queue and add the root node and its depth (1) to the queue.
  4. We enter a loop that continues until the queue is empty. In each iteration:
    • We dequeue a node and its depth (using unpacking).
    • We update our maximum depth.
    • We check if the node has children (conditional checking).
    • If it has children, we append each child and its depth (parent’s depth + 1) to the queue (updating data structures).
  5. Once the loop is finished, we return the maximum depth we’ve recorded. This integrates the ‘keeping track of depth’ drill.