LeetCode #559 — EASY

Maximum Depth of N-ary Tree

Build confidence with an intuition-first walkthrough focused on tree fundamentals.

The Problem

Problem Statement

Given a n-ary tree, find its maximum depth.

The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).

Example 1:

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

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

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.

Step 01

Brute Force Baseline

Problem summary: Given a n-ary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Tree

Example 1

[1,null,3,2,4,null,5,6]

Example 2

[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]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #559: Maximum Depth of N-ary Tree
/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/

class Solution {
    public int maxDepth(Node root) {
        if (root == null) {
            return 0;
        }
        int mx = 0;
        for (Node child : root.children) {
            mx = Math.max(mx, maxDepth(child));
        }
        return 1 + mx;
    }
}

// Accepted solution for LeetCode #559: Maximum Depth of N-ary Tree
/**
 * Definition for a Node.
 * type Node struct {
 *     Val int
 *     Children []*Node
 * }
 */

func maxDepth(root *Node) int {
	if root == nil {
		return 0
	}
	mx := 0
	for _, child := range root.Children {
		mx = max(mx, maxDepth(child))
	}
	return 1 + mx
}

# Accepted solution for LeetCode #559: Maximum Depth of N-ary Tree
"""
# 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

// Accepted solution for LeetCode #559: Maximum Depth of N-ary Tree
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #559: Maximum Depth of N-ary Tree
// /*
// // Definition for a Node.
// class Node {
//     public int val;
//     public List<Node> children;
// 
//     public Node() {}
// 
//     public Node(int _val) {
//         val = _val;
//     }
// 
//     public Node(int _val, List<Node> _children) {
//         val = _val;
//         children = _children;
//     }
// };
// */
// 
// class Solution {
//     public int maxDepth(Node root) {
//         if (root == null) {
//             return 0;
//         }
//         int mx = 0;
//         for (Node child : root.children) {
//             mx = Math.max(mx, maxDepth(child));
//         }
//         return 1 + mx;
//     }
// }

// Accepted solution for LeetCode #559: Maximum Depth of N-ary Tree
/**
 * Definition for _Node.
 * class _Node {
 *     val: number
 *     children: _Node[]
 *
 *     constructor(val?: number, children?: _Node[]) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.children = (children===undefined ? [] : children)
 *     }
 * }
 */

function maxDepth(root: _Node | null): number {
    if (!root) {
        return 0;
    }
    return 1 + Math.max(...root.children.map(child => maxDepth(child)), 0);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.