💡 Problem Formulation: Given a binary tree, the goal is to write a Python program that can find the number of nodes that are the only child of their parents. An “only child” in this context is a node that has a parent but no siblings. For example, in a binary tree where node ‘A’ has child ‘B’ (to the left) and no right child, ‘B’ would be considered an only child. The desired output is a single integer representing the count of such nodes.
Method 1: Recursive Depth-First Search
This method involves using a recursive function to traverse the binary tree using depth-first search (DFS). The function checks each node to see if it has only one child and increments a global or passed reference count variable accordingly. The time complexity of this method is O(n), where n is the number of nodes in the binary tree.
Here’s an example:
class TreeNode:
def __init__(self, value=0, left=None, right=None):
self.val = value
self.left = left
self.right = right
def count_only_child(node):
if not node:
return 0
only_child_count = 0
if bool(node.left) ^ bool(node.right): # XOR to check if only one child exists
only_child_count += 1
only_child_count += count_only_child(node.left)
only_child_count += count_only_child(node.right)
return only_child_count
# Example usage:
# A
# / \
# B C
# \
# D
# B and D are only children
root = TreeNode('A', TreeNode('B', None, TreeNode('D')), TreeNode('C'))
print(count_only_child(root))Output:
2
In the example, a simple binary tree is created with nodes ‘A’, ‘B’, ‘C’, and ‘D’, where ‘B’ and ‘D’ are only children. The count_only_child function traverses the tree recursively and returns the number of only child nodes, which in this case is 2.
Method 2: Iterative Depth-First Search
Iterative DFS uses a stack data structure to manually emulate the call stack that is naturally used in a recursive DFS solution. At each step, the current node is checked to see if it is an only child, and the count is updated accordingly. This method avoids the potential stack overflow issue that might arise in a recursive approach for very large trees.
Here’s an example:
def count_only_child_iterative(root):
if not root:
return 0
only_child_count = 0
stack = [root]
while stack:
node = stack.pop()
if bool(node.left) ^ bool(node.right):
only_child_count += 1
if node.left:
stack.append(node.left)
if node.right:
stack.append(node.right)
return only_child_count
# Using the previous tree:
print(count_only_child_iterative(root))Output:
2
The count_only_child_iterative method does exactly what the recursive version does, but it implements the depth-first search without the use of recursion using a stack. The count for only children is preserved and returned after completely traversing the tree.
Method 3: Recursive Breadth-First Search
Instead of depth-first traversal, this method applies a level-order or breadth-first search (BFS). This approach requires a queue for storing the nodes yet to be processed. The counting logic is similar to the DFS approaches—they increment the only child count when applicable. BFS might be more intuitive for some problems where processing level-wise details are important.
Here’s an example:
from collections import deque
def count_only_child_bfs(root):
if not root:
return 0
only_child_count = 0
queue = deque([root])
while queue:
node = queue.popleft()
if bool(node.left) ^ bool(node.right):
only_child_count += 1
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return only_child_count
# Using the previous tree:
print(count_only_child_bfs(root))Output:
2
The count_only_child_bfs method again computes the number of only children non-recursively, but through a queue which processes the nodes level by level. The result is the same but obtained by traversing the tree in a breadth-first manner.
Method 4: Morris In-Order Traversal
This algorithm uses a clever way to traverse the tree without additional memory for stack or queue called Morris Traversal. It modifies the tree by adding temporary links to successors, but restores the original tree as it traverses. While the Morris traversal is moving through the tree, it keeps count of the only children. This technique is space-efficient but might not be ideal for trees that shouldn’t be modified, even temporarily.
Here’s an example:
def count_only_child_morris(root):
current = root
only_child_count = 0
while current:
if current.left is None and current.right:
only_child_count += 1
current = current.right
elif current.right is None and current.left:
only_child_count += 1
current = current.left
else:
# Find the inorder predecessor of current
predecessor = current.left
while predecessor.right is not None and predecessor.right != current:
predecessor = predecessor.right
if predecessor.right is None:
# Make current the right child of its predecessor
if current.left and not current.right:
only_child_count += 1
predecessor.right = current
current = current.left
else:
# Revert the changes made to the tree
predecessor.right = None
current = current.right
return only_child_count
# Using the previous tree:
print(count_only_child_morris(root))Output:
2
The count_only_child_morris method carefully modifies the tree to carry out an in-order traversal without the use of additional memory for a stack or queue. It maintains the only child count along the way and ensures the tree structure is restored once processing is done.
Bonus One-Liner Method 5: Using Python’s functools
For those who love one-liners and Python’s functional programming features, this method uses functools.reduce to achieve the count in a single statement. It’s succinct but can be less readable to those not familiar with functional programming paradigms. Note that it’s essentially a compressed form of the recursive DFS approach.
Here’s an example:
from functools import reduce
def count_only_child_oneliner(root):
return reduce(lambda acc, n: acc + (bool(n.left) ^ bool(n.right)), [root], 0) if root else 0
# Using the previous tree:
print(count_only_child_oneliner(root))Output:
2
The count_only_child_oneliner method elegantly reduces the tree to a count of only children nodes. It succinctly expresses the counting logic but can be terse for readers not comfortable with reduce or lambda functions.
Summary/Discussion
- Method 1: Recursive DFS. Simple to understand and implement. Could lead to a stack overflow in extremely large trees.
- Method 2: Iterative DFS. Avoids stack overflow, maintains explicit control over traversal. Can be less elegant than the recursive solution.
- Method 3: Recursive BFS. Processes nodes level by level. It’s beneficial where levels of a tree hold importance. Can be memory-intense for wide trees.
- Method 4: Morris In-Order Traversal. Space efficient, as it does not use extra space for traversal stacks or queues. Not suitable for immutable tree structures due to temporary modifications.
- Method 5: Python’s functools. Clean and concise. May not be easily graspable by everyone, especially those who aren’t familiar with functional programming concepts.