#559

Easy

auto_awesomeBinary-tree traversal and state tracking

LeetCode Problem Workspace

Maximum Depth of N-ary Tree

Find the maximum depth of an N-ary tree, leveraging tree traversal techniques and state tracking.

tree depth first search breadth first search

Problem Statement

Given an N-ary tree, determine its maximum depth. The depth of the tree is the number of nodes along the longest path starting from the root node to the farthest leaf node.

The tree is represented in level-order traversal where each group of children is separated by a null value. The input structure provides a root node followed by its children, and you must compute the maximum depth of the tree from the root to the deepest leaf.

Examples

Example 1

Input: root = [1,null,3,2,4,null,5,6]

Output: 3

Example details omitted.

Example 2

Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]

Output: 5

Example details omitted.

Constraints

The total number of nodes is in the range [0, 104].
The depth of the n-ary tree is less than or equal to 1000.

Solution Approach

Depth-First Search (DFS)

Perform a DFS to explore each node's children recursively. Track the depth at each node by comparing it with the maximum depth found across its children. This approach is optimal for deep recursive trees.

Breadth-First Search (BFS)

Use BFS to explore the tree level by level. For each level, increment the depth counter, ensuring the deepest level is returned. This is effective for trees with wide breadth but shallow height.

Iterative DFS with Stack

Implement DFS iteratively using a stack to avoid recursion depth limitations. Traverse nodes while keeping track of the current depth, updating the maximum depth as you encounter deeper levels.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity for both DFS and BFS approaches is O(n), where n is the number of nodes in the tree, as each node is processed once. The space complexity for DFS is O(h), where h is the height of the tree, while for BFS, it is O(w), where w is the width of the tree (maximum number of nodes at any level).

What Interviewers Usually Probe

The candidate efficiently navigates the tree and tracks depth correctly.
The solution includes the edge cases such as an empty tree or a single-node tree.
The candidate chooses an appropriate traversal technique based on tree structure (DFS vs. BFS).

Common Pitfalls or Variants

Common pitfalls

Failing to handle edge cases like an empty tree (root = null) or single-node trees.
Not accounting for trees with varying numbers of children, leading to inaccurate depth calculations.
Incorrectly tracking depth or failing to update the maximum depth at each node.

Follow-up variants

Implement the solution with a recursive approach using DFS.
Use an iterative BFS approach to solve for the maximum depth.
Optimize the solution to handle trees with large depths or wide breadth efficiently.

FAQ

What is the best approach to solving the Maximum Depth of N-ary Tree problem?

The best approach depends on the tree structure. Use DFS for deep trees, BFS for trees with wide breadth, or an iterative DFS to avoid recursion depth issues.

How do I handle an empty tree in the Maximum Depth of N-ary Tree problem?

An empty tree can be represented by a null root. In this case, the maximum depth is 0.

What should I consider when choosing between DFS and BFS for this problem?

Choose DFS if the tree is deep and narrow, and BFS if the tree is wide. Consider memory usage and recursion depth limits as well.

Can this problem be solved using recursion?

Yes, the Maximum Depth of N-ary Tree problem is a common use case for recursion, especially using DFS to traverse and track depth.

What is the time complexity for solving the Maximum Depth of N-ary Tree problem?

The time complexity is O(n), where n is the number of nodes in the tree. Each node is visited once during traversal.

terminal

Solution

Solution 1: Recursion

First, we check if $\textit{root}$ is null. If it is, we return 0. Otherwise, we initialize a variable $\textit{mx}$ to record the maximum depth of the child nodes, then traverse all the child nodes of $\textit{root}$, recursively call the $\text{maxDepth}$ function, and update the value of $\textit{mx}$. Finally, we return $\textit{mx} + 1$.

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"""
# Definition for a Node.
class Node:
    def __init__(self, val: Optional[int] = None, children: Optional[List['Node']] = None):
        self.val = val
        self.children = children
"""


class Solution:
    def maxDepth(self, root: "Node") -> int:
        if root is None:
            return 0
        mx = 0
        for child in root.children:
            mx = max(mx, self.maxDepth(child))
        return 1 + mx

Continue Practicing

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