Leetcode #1522: Diameter of N-Ary Tree
In this guide, we solve Leetcode #1522 Diameter of N-Ary Tree in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given a root of an N-ary tree, you need to compute the length of the diameter of the tree. The diameter of an N-ary tree is the length of the longest path between any two nodes in the tree.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Tree, Depth-First Search
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
Input: root = [1,null,3,2,4,null,5,6]
Output: 3
Explanation: Diameter is shown in red color.
Python Solution
"""
# Definition for a Node.
class Node:
def __init__(self, val=None, children=None):
self.val = val
self.children = children if children is not None else []
"""
class Solution:
def diameter(self, root: 'Node') -> int:
"""
:type root: 'Node'
:rtype: int
"""
def dfs(root):
if root is None:
return 0
nonlocal ans
m1 = m2 = 0
for child in root.children:
t = dfs(child)
if t > m1:
m2, m1 = m1, t
elif t > m2:
m2 = t
ans = max(ans, m1 + m2)
return 1 + m1
ans = 0
dfs(root)
return ans
Complexity
The time complexity is O(V+E). The space complexity is O(V).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.