# Number of leaf nodes in a binary tree

Abhishek Kumar Category: Data StructuresTags: datastructures tree binary-tree

Those nodes in the tree which don't have any child are known as leaf nodes i.e., A node is a leaf node if both left and right child nodes of it are null.

To find the number of leaf nodes in a binary tree, we have to traverse each node and in the tree and check if the current node is a leaf node or not and count them one by one.

To find the number of leaf nodes in a binary tree or other trees, we have to traverse each node and in the tree and check if the current node is a leaf node or not and count them one by one.

### Leaf nodes in a binary tree

Those nodes in the tree which don't have any child are known as leaf nodes.

A node is a leaf node if both left and right child nodes of it are `null`.

So, to count all the leaf nodes we have to traverse each node in the tree and count all those nodes which are not having any child i.e, both the children are `null`.

##### Example -
```  Input:
1
/   \
2     3
/ \   / \
4   5 6   7

Output: 4
```

We can solve this problem in two ways-

### Recursive solution

For recursive solution we have to check each node one by one recursively as we have done in binary tree recursive traversal and count the nodes which are not having any child. Other types of traversal can also be used.

###### Algorithm
Leaf_Count(root)
``````1. if root is null then
2.    return 0
3. if root->left is null and root->right is null then
4.    return 1
5. else
6.    return Leaf_Count(root->left) + Leaf_Count(root->right)``````
``````class Node:
def __init__(self, data):
self.left = None
self.data = data
self.right = None

def leaf_count(root):
if root is None:
return 0
if root.left is None and root.right is None:
return 1
else:
return leaf_count(root.left) + leaf_count(root.right)

root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.left = Node(6)
root.right.right = Node(7)
print(leaf_count(root))``````
``````class Node {
constructor(data) {
this.left = null;
this.data = data;
this.right = null;
}
}

function leafCount(root) {
if (!root) return 0;
if (!root.left && !root.right) return 1;
else return leafCount(root.left) + leafCount(root.right);
}

const root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.left = new Node(6);
root.right.right = new Node(7);
console.log(leafCount(root));``````

Output

`4`

### Iterative solution

For iterative solution, we have to traverse each node in level order and count the nodes which are not having any child.

###### Algorithm
Leaf_Count(root)
``````1.  if root is null then return 0
2.  queue.enqueue(root)
3.  count = 0
4.  while queue is not empty
5.      node = queue.dequeue()
6.      if node->left is null and node->right is null then
7.          count = count + 1
8.      else
9.          if node->left is not null then
10.             queue.enqueue(node->left)
11.         if node.right is not null then
12.             queue.enqueue(node->right)
13. return count``````
``````import queue

class Node:
def __init__(self, data):
self.left = None
self.data = data
self.right = None

def leaf_count(root):
if root is None:
return 0

q = queue.Queue()
q.put(root)
count = 0
while not q.empty():
node = q.get()
if node.left is None and node.right is None:
count = count + 1
else:
if node.left:
q.put(node.left)
if node.right:
q.put(node.right)
return count

root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.left = Node(6)
root.right.right = Node(7)
print(leaf_count(root))``````
``````class Node {
constructor(data) {
this.left = null;
this.data = data;
this.right = null;
}
}

function leafCount(root) {
if (!root) return 0;

q = [];
q.push(root);
let count = 0;

while (q.length > 0) {
const node = q.shift();
if (!node.left && !node.right) count++;
else {
if (node.left) q.push(node.left);
if (node.right) q.push(node.right);
}
}
return count;
}

const root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.left = new Node(6);
root.right.right = new Node(7);
console.log(leafCount(root));``````

#### Output

Output

`4`

Time complexity: The time complexity of both of these solution will be O(n) because we are just traversing each element of node one by one.

AUTHOR

### Abhishek Kumar

Software engineer | Blogger | Keen learner
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Passionate about learning and experimenting with new things and open to more opportunities and collaborations.