Leetcode #2196: Create Binary Tree From Descriptions
In this guide, we solve Leetcode #2196 Create Binary Tree From Descriptions 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 a 2D integer array descriptions where descriptions[i] = [parenti, childi, isLefti] indicates that parenti is the parent of childi in a binary tree of unique values. Furthermore, If isLefti == 1, then childi is the left child of parenti.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Tree, Array, 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: descriptions = [[20,15,1],[20,17,0],[50,20,1],[50,80,0],[80,19,1]]
Output: [50,20,80,15,17,19]
Explanation: The root node is the node with value 50 since it has no parent.
The resulting binary tree is shown in the diagram.
Python Solution
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def createBinaryTree(self, descriptions: List[List[int]]) -> Optional[TreeNode]:
nodes = defaultdict(TreeNode)
children = set()
for parent, child, isLeft in descriptions:
if parent not in nodes:
nodes[parent] = TreeNode(parent)
if child not in nodes:
nodes[child] = TreeNode(child)
children.add(child)
if isLeft:
nodes[parent].left = nodes[child]
else:
nodes[parent].right = nodes[child]
root = (set(nodes.keys()) - children).pop()
return nodes[root]
Complexity
The time complexity is , and the space complexity is , where is the length of . The space complexity is , where is the length of .
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.