Leetcode #1361: Validate Binary Tree Nodes
In this guide, we solve Leetcode #1361 Validate Binary Tree Nodes 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 have n binary tree nodes numbered from 0 to n - 1 where node i has two children leftChild[i] and rightChild[i], return true if and only if all the given nodes form exactly one valid binary tree. If node i has no left child then leftChild[i] will equal -1, similarly for the right child.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Tree, Depth-First Search, Breadth-First Search, Union Find, Graph, Binary Tree
Intuition
The data forms a graph, so we should explore nodes and edges systematically.
A traversal ensures we visit each node once while maintaining the needed state.
Approach
Build an adjacency list and traverse with BFS or DFS.
Aggregate results as you visit nodes.
Steps:
- Build the graph.
- Traverse with BFS/DFS.
- Accumulate the required output.
Example
Input: n = 4, leftChild = [1,-1,3,-1], rightChild = [2,-1,-1,-1]
Output: true
Python Solution
class Solution:
def validateBinaryTreeNodes(
self, n: int, leftChild: List[int], rightChild: List[int]
) -> bool:
def find(x: int) -> int:
if p[x] != x:
p[x] = find(p[x])
return p[x]
p = list(range(n))
vis = [False] * n
for i, (a, b) in enumerate(zip(leftChild, rightChild)):
for j in (a, b):
if j != -1:
if vis[j] or find(i) == find(j):
return False
p[find(i)] = find(j)
vis[j] = True
n -= 1
return n == 1
Complexity
The time complexity is , and the space complexity is . The space complexity is .
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.