Leetcode #1485: Clone Binary Tree With Random Pointer
In this guide, we solve Leetcode #1485 Clone Binary Tree With Random Pointer 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
A binary tree is given such that each node contains an additional random pointer which could point to any node in the tree or null. Return a deep copy of the tree.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Tree, Depth-First Search, Breadth-First Search, Hash Table, Binary Tree
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: root = [[1,null],null,[4,3],[7,0]]
Output: [[1,null],null,[4,3],[7,0]]
Explanation: The original binary tree is [1,null,4,7].
The random pointer of node one is null, so it is represented as [1, null].
The random pointer of node 4 is node 7, so it is represented as [4, 3] where 3 is the index of node 7 in the array representing the tree.
The random pointer of node 7 is node 1, so it is represented as [7, 0] where 0 is the index of node 1 in the array representing the tree.
Python Solution
# Definition for Node.
# class Node:
# def __init__(self, val=0, left=None, right=None, random=None):
# self.val = val
# self.left = left
# self.right = right
# self.random = random
class Solution:
def copyRandomBinaryTree(self, root: 'Optional[Node]') -> 'Optional[NodeCopy]':
def dfs(root):
if root is None:
return None
if root in mp:
return mp[root]
copy = NodeCopy(root.val)
mp[root] = copy
copy.left = dfs(root.left)
copy.right = dfs(root.right)
copy.random = dfs(root.random)
return copy
mp = {}
return dfs(root)
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.