Given A binary Tree, how do you count all the half nodes (which has only one child) without using recursion? Note leaves should not be touched as they have both children as NULL.
Input : Root of below tree
Output : 3 Nodes 7, 5 and 9 are half nodes as one of their child is Null. So count of half nodes in the above tree is 3
Iterative
The idea is to use level-order traversal to solve this problem efficiently.
1) Create an empty Queue Node and push root node to Queue.
2) Do following while nodeQeue is not empty.
a) Pop an item from Queue and process it.
a.1) If it is half node then increment count++.
b) Push left child of popped item to Queue, if available.
c) Push right child of popped item to Queue, if available.Below is the implementation of this idea.
C++
// C++ program to count half nodes in a Binary Tree #include <bits/stdc++.h> using namespace std;
// A binary tree Node has data, pointer to left // child and a pointer to right child struct Node
{ int data;
struct Node* left, *right;
}; // Function to get the count of half Nodes in // a binary tree unsigned int gethalfCount(struct Node* node)
{ // If tree is empty
if (!node)
return 0;
int count = 0; // Initialize count of half nodes
// Do level order traversal starting from root
queue<Node *> q;
q.push(node);
while (!q.empty())
{
struct Node *temp = q.front();
q.pop();
if (!temp->left && temp->right ||
temp->left && !temp->right)
count++;
if (temp->left != NULL)
q.push(temp->left);
if (temp->right != NULL)
q.push(temp->right);
}
return count;
} /* Helper function that allocates a new Node with the given data and NULL left
and right pointers. */
struct Node* newNode(int data)
{ struct Node* node = new Node;
node->data = data;
node->left = node->right = NULL;
return (node);
} // Driver program int main(void)
{ /* 2
/ \
7 5
\ \
6 9
/ \ /
1 11 4
Let us create Binary Tree shown in
above example */
struct Node *root = newNode(2);
root->left = newNode(7);
root->right = newNode(5);
root->left->right = newNode(6);
root->left->right->left = newNode(1);
root->left->right->right = newNode(11);
root->right->right = newNode(9);
root->right->right->left = newNode(4);
cout << gethalfCount(root);
return 0;
} |
Java
// Java program to count half nodes in a Binary Tree // using Iterative approach import java.util.Queue;
import java.util.LinkedList;
// Class to represent Tree node class Node
{ int data;
Node left, right;
public Node(int item)
{
data = item;
left = null;
right = null;
}
} // Class to count half nodes of Tree class BinaryTree
{ Node root;
/* Function to get the count of half Nodes in
a binary tree*/
int gethalfCount()
{
// If tree is empty
if (root==null)
return 0;
// Do level order traversal starting from root
Queue<Node> queue = new LinkedList<Node>();
queue.add(root);
int count=0; // Initialize count of half nodes
while (!queue.isEmpty())
{
Node temp = queue.poll();
if (temp.left!=null && temp.right==null ||
temp.left==null && temp.right!=null)
count++;
// Enqueue left child
if (temp.left != null)
queue.add(temp.left);
// Enqueue right child
if (temp.right != null)
queue.add(temp.right);
}
return count;
}
public static void main(String args[])
{
/* 2
/ \
7 5
\ \
6 9
/ \ /
1 11 4
Let us create Binary Tree shown in
above example */
BinaryTree tree_level = new BinaryTree();
tree_level.root = new Node(2);
tree_level.root.left = new Node(7);
tree_level.root.right = new Node(5);
tree_level.root.left.right = new Node(6);
tree_level.root.left.right.left = new Node(1);
tree_level.root.left.right.right = new Node(11);
tree_level.root.right.right = new Node(9);
tree_level.root.right.right.left = new Node(4);
System.out.println(tree_level.gethalfCount());
}
} |
Python3
# Python program to count # half nodes in a Binary Tree # using iterative approach # A node structure class Node:
# A utility function to create a new node
def __init__(self ,key):
self.data = key
self.left = None
self.right = None
# Iterative Method to count half nodes of binary tree def gethalfCount(root):
# Base Case
if root is None:
return 0
# Create an empty queue for level order traversal
queue = []
# Enqueue Root and initialize count
queue.append(root)
count = 0 #initialize count for half nodes
while(len(queue) > 0):
node = queue.pop(0)
# if it is half node then increment count
if node.left is not None and node.right is None or node.left is None and node.right is not None:
count = count+1
#Enqueue left child
if node.left is not None:
queue.append(node.left)
# Enqueue right child
if node.right is not None:
queue.append(node.right)
return count
#Driver Program to test above function root = Node(2)
root.left = Node(7)
root.right = Node(5)
root.left.right = Node(6)
root.left.right.left = Node(1)
root.left.right.right = Node(11)
root.right.right = Node(9)
root.right.right.left = Node(4)
print "%d" %(gethalfCount(root))
|
C#
// C# program to count half nodes in a Binary Tree // using Iterative approach using System;
using System.Collections.Generic;
// Class to represent Tree node public class Node
{ public int data;
public Node left, right;
public Node(int item)
{
data = item;
left = null;
right = null;
}
} // Class to count half nodes of Tree public class BinaryTree
{ Node root;
/* Function to get the count of half Nodes in
a binary tree*/
int gethalfCount()
{
// If tree is empty
if (root == null)
return 0;
// Do level order traversal starting from root
Queue<Node> queue = new Queue<Node>();
queue.Enqueue(root);
int count = 0; // Initialize count of half nodes
while (queue.Count != 0)
{
Node temp = queue.Dequeue();
if (temp.left != null && temp.right == null ||
temp.left == null && temp.right != null)
count++;
// Enqueue left child
if (temp.left != null)
queue.Enqueue(temp.left);
// Enqueue right child
if (temp.right != null)
queue.Enqueue(temp.right);
}
return count;
}
// Driver code
public static void Main()
{
/* 2
/ \
7 5
\ \
6 9
/ \ /
1 11 4
Let us create Binary Tree shown in
above example */
BinaryTree tree_level = new BinaryTree();
tree_level.root = new Node(2);
tree_level.root.left = new Node(7);
tree_level.root.right = new Node(5);
tree_level.root.left.right = new Node(6);
tree_level.root.left.right.left = new Node(1);
tree_level.root.left.right.right = new Node(11);
tree_level.root.right.right = new Node(9);
tree_level.root.right.right.left = new Node(4);
Console.WriteLine(tree_level.gethalfCount());
}
} // This code contributed by Rajput-Ji |
Javascript
<script> // Javascript program to count half nodes in a Binary Tree // using Iterative approach // Class to represent Tree node class Node { constructor(item)
{
this.data = item;
this.left = null;
this.right = null;
}
} // Class to count half nodes of Tree var root;
/* Function to get the count of half Nodes in a binary tree*/function gethalfCount()
{ // If tree is empty
if (root == null)
return 0;
// Do level order traversal starting from root
var queue = [];
queue.push(root);
var count = 0; // Initialize count of half nodes
while (queue.length != 0)
{
var temp = queue.shift();
if (temp.left != null && temp.right == null ||
temp.left == null && temp.right != null)
count++;
// push left child
if (temp.left != null)
queue.push(temp.left);
// push right child
if (temp.right != null)
queue.push(temp.right);
}
return count;
} // Driver code /* 2 / \ 7 5 \ \ 6 9 / \ / 1 11 4 Let us create Binary Tree shown in above example */root = new Node(2);
root.left = new Node(7);
root.right = new Node(5);
root.left.right = new Node(6);
root.left.right.left = new Node(1);
root.left.right.right = new Node(11);
root.right.right = new Node(9);
root.right.right.left = new Node(4);
document.write(gethalfCount()); // This code is contributed by famously. </script> |
Output:
3
Time Complexity: O(n)
Auxiliary Space: O(n)
where, n is number of nodes in given binary tree
Recursive
The idea is to traverse the tree in postorder. If current node is half, we increment result by 1 and add returned values of left and right subtrees.
C++
// C++ program to count half nodes in a Binary Tree #include <bits/stdc++.h> using namespace std;
// A binary tree Node has data, pointer to left // child and a pointer to right child struct Node
{ int data;
struct Node* left, *right;
}; // Function to get the count of half Nodes in // a binary tree unsigned int gethalfCount(struct Node* root)
{ if (root == NULL)
return 0;
int res = 0;
if ((root->left == NULL && root->right != NULL) ||
(root->left != NULL && root->right == NULL))
res++;
res += (gethalfCount(root->left) + gethalfCount(root->right));
return res;
} /* Helper function that allocates a new Node with the given data and NULL left
and right pointers. */
struct Node* newNode(int data)
{ struct Node* node = new Node;
node->data = data;
node->left = node->right = NULL;
return (node);
} // Driver program int main(void)
{ /* 2
/ \
7 5
\ \
6 9
/ \ /
1 11 4
Let us create Binary Tree shown in
above example */
struct Node *root = newNode(2);
root->left = newNode(7);
root->right = newNode(5);
root->left->right = newNode(6);
root->left->right->left = newNode(1);
root->left->right->right = newNode(11);
root->right->right = newNode(9);
root->right->right->left = newNode(4);
cout << gethalfCount(root);
return 0;
} |
Java
// Java program to count half nodes in a Binary Tree import java.util.*;
class GfG {
// A binary tree Node has data, pointer to left // child and a pointer to right child static class Node
{ int data;
Node left, right;
} // Function to get the count of half Nodes in // a binary tree static int gethalfCount(Node root)
{ if (root == null)
return 0;
int res = 0;
if ((root.left == null && root.right != null) ||
(root.left != null && root.right == null))
res++;
res += (gethalfCount(root.left)
+ gethalfCount(root.right));
return res;
} /* Helper function that allocates a new Node with the given data and NULL left and right pointers. */static Node newNode(int data)
{ Node node = new Node();
node.data = data;
node.left = null;
node.right = null;
return (node);
} // Driver program public static void main(String[] args)
{ /* 2
/ \
7 5
\ \
6 9
/ \ /
1 11 4
Let us create Binary Tree shown in
above example */
Node root = newNode(2);
root.left = newNode(7);
root.right = newNode(5);
root.left.right = newNode(6);
root.left.right.left = newNode(1);
root.left.right.right = newNode(11);
root.right.right = newNode(9);
root.right.right.left = newNode(4);
System.out.println(gethalfCount(root));
} } |
Python3
# Python program to count half nodes in a binary tree # A node structure class newNode:
def __init__(self, data):
self.data = data
self.left = self.right = None
# Function to get the count of half Nodes in a binary tree def gethalfCount(root):
if root == None:
return 0
res = 0
if(root.left == None and root.right != None) or \
(root.left != None and root.right == None):
res += 1
res += (gethalfCount(root.left) + \
gethalfCount(root.right))
return res
# Driver program ''' 2 / \
7 5
\ \
6 9
/ \ /
1 11 4
Let us create Binary Tree shown in
above example '''
root = newNode(2)
root.left = newNode(7)
root.right = newNode(5)
root.left.right = newNode(6)
root.left.right.left = newNode(1)
root.left.right.right = newNode(11)
root.right.right = newNode(9)
root.right.right.left = newNode(4)
print(gethalfCount(root))
# This code is contributed by simranjenny84 |
C#
// C# program to count half nodes in a Binary Tree using System;
class GfG
{ // A binary tree Node has data, pointer to left
// child and a pointer to right child
public class Node
{
public int data;
public Node left, right;
}
// Function to get the count of half Nodes in
// a binary tree
static int gethalfCount(Node root)
{
if (root == null)
return 0;
int res = 0;
if ((root.left == null && root.right != null) ||
(root.left != null && root.right == null))
res++;
res += (gethalfCount(root.left)
+ gethalfCount(root.right));
return res;
}
/* Helper function that allocates a new
Node with the given data and NULL left
and right pointers. */
static Node newNode(int data)
{
Node node = new Node();
node.data = data;
node.left = null;
node.right = null;
return (node);
}
// Driver code
public static void Main()
{
/* 2
/ \
7 5
\ \
6 9
/ \ /
1 11 4
Let us create Binary Tree shown in
above example */
Node root = newNode(2);
root.left = newNode(7);
root.right = newNode(5);
root.left.right = newNode(6);
root.left.right.left = newNode(1);
root.left.right.right = newNode(11);
root.right.right = newNode(9);
root.right.right.left = newNode(4);
Console.WriteLine(gethalfCount(root));
}
} /* This code contributed by PrinciRaj1992 */ |
Javascript
<script> // Javascript program to count half // nodes in a Binary Tree class Node { constructor(data)
{
this.left = null;
this.right = null;
this.data = data;
}
} // Function to get the count of half // Nodes in a binary tree function gethalfCount(root)
{ if (root == null)
return 0;
let res = 0;
if ((root.left == null && root.right != null) ||
(root.left != null && root.right == null))
res++;
res += (gethalfCount(root.left) +
gethalfCount(root.right));
return res;
} /* Helper function that allocates a new Node with the given data and NULL left and right pointers. */function newNode(data)
{ let node = new Node(data);
return (node);
} // Driver code /* 2 / \
7 5
\ \
6 9
/ \ /
1 11 4
Let us create Binary Tree shown in above example */let root = newNode(2); root.left = newNode(7); root.right = newNode(5); root.left.right = newNode(6); root.left.right.left = newNode(1); root.left.right.right = newNode(11); root.right.right = newNode(9); root.right.right.left = newNode(4); document.write(gethalfCount(root)); // This code is contributed by divyeshrabadiya07 </script> |
Output :
3
Time Complexity: O(n)
Auxiliary Space: O(n)
where, n is number of nodes in given binary tree
Similar Articles:
- Program to get count of leaf nodes in Binary Tree
- Given a binary tree, how do you remove all the half nodes?
This article is contributed by Mr. Somesh Awasthi and Rakesh Kumar.