Tree Traversals (Counting)

Problem Overview

Learning Goals

After solving this problem, you will be able to:

• Understand how tree structure constrains traversal orderings
• Apply Catalan numbers to count valid binary tree configurations
• Use factorial counting for sibling permutations in rooted trees
• Implement modular arithmetic for large counting results

Problem Statement

Problem: Given a rooted tree structure, count the number of distinct traversal orderings possible, or given traversal sequences, count how many unique trees produce them.

Input:

• Line 1: Integer n (number of nodes)
• Following lines: Tree structure (edges or parent array)

Output:

• Number of distinct traversal orderings modulo 10^9 + 7

Constraints:

• 1 <= n <= 2 * 10^5

Example

Input:
5
1 1 2 2

Tree Structure:
       1
      / \
     2   3
    / \
   4   5

Output:
4

Explanation: Node 1 has 2 children (2, 3) that can be visited in 2! = 2 orders. Node 2 has 2 children (4, 5) that can be visited in 2! = 2 orders. Total: 2 * 2 = 4 distinct traversals.

Intuition: How to Think About This Problem

Pattern Recognition

Key Question: What determines the number of valid traversal orderings for a tree?

The number of traversal orderings depends on how many ways we can order the children of each node. For each node with k children, there are k! ways to arrange them.

Breaking Down the Problem

1. What are we looking for? Total distinct DFS orderings of the tree
2. What information do we have? Tree structure with parent-child relationships
3. What’s the relationship? Product of factorials of children counts

Key Insight

For a rooted tree where children can be visited in any order:

Total orderings = Product of (children_count[v])! for all nodes v

Visual Example

       1 (2 children)
      / \
     2   3 (0 children)
    / \
   4   5 (0 children each)

Node 1: 2 children -> 2! = 2 orderings
Node 2: 2 children -> 2! = 2 orderings
Node 3: 0 children -> 0! = 1 ordering
Total: 2 * 2 * 1 = 4 orderings

Solution 1: Brute Force (Enumeration)

Idea

Generate all possible traversal orderings by trying every permutation of children at each node.

Algorithm

1. Build adjacency list from parent array
2. For each node, generate all permutations of children
3. Recursively count all valid orderings

Code

from itertools import permutations

def count_traversals_brute(n, parents):
  """
  Brute force: enumerate all orderings.

  Time: O(n! * n) in worst case
  Space: O(n)
  """
  # Build adjacency list
  children = [[] for _ in range(n + 1)]
  for i, p in enumerate(parents, start=2):
    children[p].append(i)

  def count_orderings(node):
    if not children[node]:
      return 1

    total = 0
    for perm in permutations(children[node]):
      product = 1
      for child in perm:
        product *= count_orderings(child)
      total += product
    return total

  return count_orderings(1)

Complexity

Why This Works (But Is Slow)

Correctly enumerates all orderings but factorial growth makes it impractical for n > 10.

Solution 2: Optimal (Factorial Product)

Key Insight

The Trick: Instead of enumerating, multiply factorials of children counts.

For each node, the number of ways to order its children is (number of children)!. The total orderings is the product across all nodes.

Formula

answer = Product of factorial(children_count[v]) for all v
       = factorial(c1) * factorial(c2) * ... * factorial(cn)

Algorithm

1. Build tree from parent array
2. Count children for each node
3. Compute product of factorials (with modular arithmetic)

Dry Run Example

Input: n=5, parents=[1, 1, 2, 2]

Step 1: Build children lists
  Node 1: children = [2, 3]  (count = 2)
  Node 2: children = [4, 5]  (count = 2)
  Node 3: children = []      (count = 0)
  Node 4: children = []      (count = 0)
  Node 5: children = []      (count = 0)

Step 2: Compute factorials
  fact[0] = 1
  fact[1] = 1
  fact[2] = 2

Step 3: Multiply
  result = fact[2] * fact[2] * fact[0] * fact[0] * fact[0]
         = 2 * 2 * 1 * 1 * 1
         = 4

Output: 4

Code (Python)

def count_traversals(n, parents):
  """
  Count tree traversal orderings using factorial product.

  Time: O(n)
  Space: O(n)
  """
  MOD = 10**9 + 7

  # Precompute factorials
  fact = [1] * (n + 1)
  for i in range(1, n + 1):
    fact[i] = fact[i-1] * i % MOD

  # Count children for each node
  children_count = [0] * (n + 1)
  for i, p in enumerate(parents, start=2):
    children_count[p] += 1

  # Compute product of factorials
  result = 1
  for v in range(1, n + 1):
    result = result * fact[children_count[v]] % MOD

  return result


def solve():
  n = int(input())
  if n == 1:
    print(1)
    return
  parents = list(map(int, input().split()))
  print(count_traversals(n, parents))

Complexity

Common Mistakes

Mistake 1: Forgetting Modular Arithmetic

# WRONG - Integer overflow for large n
result = 1
for v in range(1, n + 1):
  result *= factorial(children_count[v])  # Overflows!

# CORRECT - Apply modulo at each step
result = 1
for v in range(1, n + 1):
  result = result * fact[children_count[v]] % MOD

Problem: Factorials grow extremely fast; 21! exceeds 64-bit integers. Fix: Always apply modulo after each multiplication.

Mistake 2: Off-by-One in Parent Indexing

# WRONG - Parents are for nodes 2..n, not 1..n
for i, p in enumerate(parents):  # i starts at 0
  children_count[p] += 1

# CORRECT - Parents start at node 2
for i, p in enumerate(parents, start=2):
  children_count[p] += 1

Problem: Node 1 is the root with no parent in input. Fix: Enumerate from 2 when processing parent array.

Mistake 3: Not Handling Single Node Case

# WRONG - Empty parent array causes error
parents = list(map(int, input().split()))  # Empty for n=1

# CORRECT - Handle n=1 separately
if n == 1:
  print(1)
  return

Edge Cases

When to Use This Pattern

Use This Approach When:

• Counting orderings of tree traversals
• Computing permutations with tree constraints
• Problems involving child ordering independence

Don’t Use When:

• Tree structure is constrained (e.g., BST with ordering)
• Specific traversal order required (preorder/inorder/postorder)
• Need actual traversal sequences, not just count

Pattern Recognition Checklist:

• Each node’s children can be visited in any order? -> Factorial product
• Counting binary trees from traversal? -> Catalan numbers
• Counting labeled trees? -> Cayley’s formula (n^(n-2))

Related Problems

Easier (Do These First)

Similar Difficulty

Harder (Do These After)

Key Takeaways

1. The Core Idea: Tree traversal orderings = product of factorials of children counts
2. Time Optimization: O(n!) enumeration -> O(n) factorial product
3. Space Trade-off: O(n) for factorial precomputation
4. Pattern: Combinatorial counting with tree structure constraints

Practice Checklist

Before moving on, make sure you can:

• Derive the factorial product formula from first principles
• Implement modular factorial computation
• Handle edge cases (n=1, linear chain, star graph)
• Explain why children orderings are independent

Additional Resources

• CP-Algorithms: Catalan Numbers
• CSES Problem Set - Tree Algorithms
• Modular Arithmetic Tutorial