Leetcode #1506: Find Root of N-Ary Tree
In this guide, we solve Leetcode #1506 Find Root 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
You are given all the nodes of an N-ary tree as an array of Node objects, where each node has a unique value. Return the root of the N-ary tree.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Bit Manipulation, Tree, Depth-First Search, Hash Table
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: tree = [1,null,3,2,4,null,5,6]
Output: [1,null,3,2,4,null,5,6]
Explanation: The tree from the input data is shown above.
The driver code creates the tree and gives findRoot the Node objects in an arbitrary order.
For example, the passed array could be [Node(5),Node(4),Node(3),Node(6),Node(2),Node(1)] or [Node(2),Node(6),Node(1),Node(3),Node(5),Node(4)].
The findRoot function should return the root Node(1), and the driver code will serialize it and compare with the input data.
The input data and serialized Node(1) are the same, so the test passes.
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 findRoot(self, tree: List['Node']) -> 'Node':
x = 0
for node in tree:
x ^= node.val
for child in node.children:
x ^= child.val
return next(node for node in tree if node.val == x)
Complexity
The time complexity is O(n). The space complexity is O(n).
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.