Python – Leaf and Non-Leaf Nodes Dictionary

Sometimes, while working with Python, we can have a problem in which we need to work with Graph data represented in form of a Dictionary. In this, we may need to check for all the leaf and non-lead nodes. This kind of problem has direct applications in algorithms of Machine Learning. Let’s discuss a way in which this task can be performed.

Input : test_dict = {‘Gfg’: {‘is’: 2, ‘best’: 1}} 
Output : {‘leaf’: 2, ‘non-leaf’: 1}

Input : test_dict = {‘Gfg’: {‘best’: 2}} 
Output : {‘non-leaf’: 1, ‘leaf’: 1} 

Method 1: Using recursion + isinstance() + loop
The combination of the above functionalities can be used to solve this problem. In this, we recur for the inner nesting using recursion and check for leaves or non-leaves by value types using isinstance().

Python3

# Python3 code to demonstrate working of
# Leaf and Non-Leaf Nodes Dictionary
# Using recursion + isinstance() + loop

def hlpr_fnc(dict1, res = {'non-leaf':0, 'leaf':0}):

    #res['non-leaf'] += 1
    nodes = dict1.keys()
    for node in nodes:
      subnode = dict1[node]
      if isinstance(subnode, dict):
        res['non-leaf'] += 1
        hlpr_fnc(subnode, res)
      else:
        res['leaf'] += 1
    return res

# initializing dictionary
test_dict = {'a': {'b': 1, 'c': {'d': {'e': 2, 'f': 1}}, 'g': {'h': {'i': 2, 'j': 1}}}}

# printing original dictionary
print("The original dictionary : " + str(test_dict))

# Leaf and Non-Leaf Nodes Dictionary
# Using recursion + isinstance() + loop
res = hlpr_fnc(test_dict)

# printing result
print("The leaf and Non-Leaf nodes : " + str(res))

Output:

The original dictionary : {'a': {'b': 1, 'c': {'d': {'e': 2, 'f': 1}}, 'g': {'h': {'i': 2, 'j': 1}}}}
The leaf and Non-Leaf nodes : {'non-leaf': 5, 'leaf': 5}

Time Complexity: O(n), where n is the number of nodes in the dictionary.

Auxiliary Space: O(n)

Method 2 : Using stack and loop

• Initialize two variables, non_leaf and leaf, to zero.
• Create an empty list stack and append the input dictionary to it.
• Use a while loop to iterate over the stack until it is empty.
• Within the loop, pop the last dictionary from the stack.
• For each key-value pair in the popped dictionary, increment non_leaf by 1 if the value is a dictionary and append it to the stack. Otherwise, increment leaf by 1.
• Return a dictionary with non-leaf and leaf as keys and their respective counts as values.

Python3

def leaf_and_nonleaf_nodes(dict1):
    non_leaf = 0
    leaf = 0
    stack = [dict1]
    while stack:
        curr_dict = stack.pop()
        for val in curr_dict.values():
            if isinstance(val, dict):
                non_leaf += 1
                stack.append(val)
            else:
                leaf += 1
    return {'non-leaf': non_leaf, 'leaf': leaf}

# initializing dictionary
test_dict = {'a': {'b': 1, 'c': {'d': {'e': 2, 'f': 1}}, 'g': {'h': {'i': 2, 'j': 1}}}}

# printing original dictionary
print("The original dictionary : " + str(test_dict))

# Leaf and Non-Leaf Nodes Dictionary
# Using stack and loop
res = leaf_and_nonleaf_nodes(test_dict)

# printing result
print("The leaf and Non-Leaf nodes : " + str(res))

Output

The original dictionary : {'a': {'b': 1, 'c': {'d': {'e': 2, 'f': 1}}, 'g': {'h': {'i': 2, 'j': 1}}}}
The leaf and Non-Leaf nodes : {'non-leaf': 5, 'leaf': 5}

Time complexity: O(N), where N is the number of nodes in the input dictionary.
Auxiliary space: O(N), where N is the maximum number of nodes stored in the stack at any given time.

Method 3 :Using a queue:

Algorithm:

1. Initialize two variables non_leaf and leaf to 0.
2. Create a deque and append the input dictionary to it.
3. While the deque is not empty, perform the following steps:
   a. Pop the leftmost element from the deque and assign it to the curr_dict variable.
   b. For each value in curr_dict, do the following:
   i. Check if the value is a dictionary.
   ii. If the value is a dictionary, increment the non_leaf variable and append the value to the deque.
   iii. If the value is not a dictionary, increment the leaf variable.
4. Create a dictionary with the non-leaf and leaf variables as values and return it.

Python3

from collections import deque

def leaf_and_nonleaf_nodes_queue(dict1):
    non_leaf = 0
    leaf = 0
    queue = deque([dict1])
    while queue:
        curr_dict = queue.popleft()
        for val in curr_dict.values():
            if isinstance(val, dict):
                non_leaf += 1
                queue.append(val)
            else:
                leaf += 1
    return {'non-leaf': non_leaf, 'leaf': leaf}

# initializing dictionary
test_dict = {'a': {'b': 1, 'c': {'d': {'e': 2, 'f': 1}}, 'g': {'h': {'i': 2, 'j': 1}}}}

# printing original dictionary
print("The original dictionary : " + str(test_dict))

# Leaf and Non-Leaf Nodes Dictionary
# Using queue
res = leaf_and_nonleaf_nodes_queue(test_dict)

# printing result
print("The leaf and Non-Leaf nodes : " + str(res))
#This code is contributed by Jyothi pinjala

Output

The original dictionary : {'a': {'b': 1, 'c': {'d': {'e': 2, 'f': 1}}, 'g': {'h': {'i': 2, 'j': 1}}}}
The leaf and Non-Leaf nodes : {'non-leaf': 5, 'leaf': 5}

Time Complexity:  O(N), where N is the total number of elements in the input dictionary. This is because each element in the dictionary is visited only once.

Auxiliary Space:  O(N), where N is the total number of elements in the input dictionary. This is because the deque used to store the dictionary elements can potentially store all N elements.