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.

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`

.

Input: 1 / \ 2 3 / \ / \ 4 5 6 7Output: 4

We can solve this problem in two ways-

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.

__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

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

__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

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__