5 Best Ways to Python Program to Count Number of Leaf Nodes in a Tree

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πŸ’‘ Problem Formulation: In computational trees, leaf nodes are those without children, resembling the leaves on a tree. Calculating the number of leaf nodes can be crucial for understanding tree complexity or optimizing traversal. Our goal is to write a Python program that counts the number of leaf nodes in a given binary tree. For example, given a tree object, our function should return the number of leaf nodes as an integer.

Method 1: Recursive Approach

This method leverages the inherent recursive nature of trees. Every tree node is potentially the root of a sub-tree. The function recursively traverses each node, counting the leaves as it goes, which are identified as nodes that have no children.

Here’s an example:

class TreeNode:
    def __init__(self, value):
        self.value = value
        self.left = None
        self.right = None

def count_leaves(node):
    if node is None:
        return 0
    if not node.left and not node.right:
        return 1
    return count_leaves(node.left) + count_leaves(node.right)

# Example Tree
root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
root.left.left = TreeNode(4)
root.left.right = TreeNode(5)


The output of this code snippet is 3.

This snippet defines a simple tree and uses a recursive function, count_leaves(), to count leaf nodes. If a node is null, it contributes no leaves. If a node has no children, it is a leaf node and counts for one. Otherwise, we add up the leaf nodes from the left and right subtrees.

Method 2: Iterative Deep-First Search (DFS)

DFS traverses the tree depth-wise, using a stack to keep track of nodes. For every node processed, if it is a leaf, increment the count. This method effectively emulates the call stack used in recursion, iteratively.

Here’s an example:

def count_leaves_iterative(root):
    if not root:
        return 0
    stack, leaf_count = [root], 0
    while stack:
        node = stack.pop()
        if node.left:
        if node.right:
        if not node.left and not node.right:
            leaf_count += 1
    return leaf_count


The output of this code snippet is 3.

This code uses a stack to perform DFS and count the leaf nodes iteratively. It examines each node to determine if it is a leaf by checking the absence of both left and right children. This approach can be beneficial when avoiding the potential stack overflow of recursive methods.

Method 3: Iterative Breadth-First Search (BFS)

BFS explores the tree level by level using a queue. This method engages in a broader search, spreading outwards and examining all nodes at each depth before going deeper. A node is considered a leaf if it has no children.

Here’s an example:

from collections import deque

def count_leaves_bfs(root):
    if not root:
        return 0
    queue, leaf_count = deque([root]), 0
    while queue:
        node = queue.popleft()
        if not node.left and not node.right:
            leaf_count += 1
        if node.left:
        if node.right:
    return leaf_count


The output is again 3.

This snippet demonstrates the use of a queue to perform a BFS on the tree. The nodes are examined in breadth-wise order, and leaves are counted when neither left nor right children are present. This method can be more memory-efficient compared to DFS for wide trees.

Method 4: Using Morris Traversal

Morris Traversal is a space-efficient tree traversal algorithm that leverages the tree’s structure itself to keep track of the traversal by creating temporary links in the tree. It traverses the tree without recursion and without a stack, thus saving space.

Here’s an example:

def count_leaves_morris(root):
    current, leaf_count = root, 0
    while current:
        if not current.left and not current.right:
            leaf_count += 1
        if current.left:
            current = current.left
        elif current.right:
            current = current.right
            # Restore the tree to its original state if modified during traversal
            # and move to the next node
    return leaf_count

# Won't run Morris Traversal on the example tree as it requires modification of the tree

The code snippet does not have an output and is a conceptual outline, as the Morris Traversal needs to handle the restoration of the tree (not implemented in the code).

This code gives a conceptual outline of using the Morris traversal for counting leaf nodes without altering the example tree. It saves memory by not using additional data structures but requires careful handling to ensure the original tree structure is preserved.

Bonus One-Liner Method 5: Using Higher-Order Functions

If the tree structure supports iterators or can be converted into a list easily, Python’s higher-order functions like filter() and map() can be utilized to count leaf nodes succinctly. However, this is not traditionally applicable in tree data structures and may require a custom tree to list conversion method.

Here’s an example:

# Assuming we have a function that returns all the nodes in a tree as a list
# all_nodes = get_all_nodes_as_list(root)
# leaf_count = len(list(filter(lambda node: not node.left and not node.right, all_nodes)))

If get_all_nodes_as_list existed and returned all nodes, the output would depend on the tree provided.

The presumed one-liner uses filter() to sift through all nodes, picking out leaf nodes and counting them with len(). This method hinges on an efficient way to linearize the tree into a list, which is often not practical, hence its status as a conceptual ‘bonus’ method.


  • Method 1: Recursive Approach. Intuitive for tree structures. Stack overflow risk for deep trees.
  • Method 2: Iterative DFS. Avoids potential stack overflows. Slightly complex logic using stack explicitly.
  • Method 3: Iterative BFS. Memory efficient for wide trees. More nodes kept in memory at once.
  • Method 4: Using Morris Traversal. Space efficient. Complex logic and potential to disrupt tree structure.
  • Bonus Method 5: Higher-Order Functions. Elegant one-liner. Impractical for traditional tree structures.