Given A binary Tree, how do you count all the full nodes (Nodes which have both children as not NULL) without using recursion and with recursion? Note leaves should not be touched as they have both children as NULL.
Nodes 2 and 6 are full nodes has both child’s. So count of full nodes in the above tree is 2
Method: 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 full 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 // full 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 full Nodes in // a binary tree unsigned int getfullCount(struct Node* node) { // If tree is empty if (!node) return 0; queue<Node *> q; // Do level order traversal starting from root int count = 0; // Initialize count of full nodes q.push(node); while (!q.empty()) { struct Node *temp = q.front(); q.pop(); if (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 as shown */ 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 << getfullCount(root); return 0; } |
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Java
// Java program to count // full 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 full nodes of Tree class BinaryTree { Node root; /* Function to get the count of full Nodes in a binary tree*/ int getfullCount() { // If tree is empty if (root==null) return 0; // Initialize empty queue. Queue<Node> queue = new LinkedList<Node>(); // Do level order traversal starting from root queue.add(root); int count=0; // Initialize count of full nodes while (!queue.isEmpty()) { Node temp = queue.poll(); if (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 as shown */ 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.getfullCount()); } } |
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Python
# Python program to count # full 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 full nodes of binary tree def getfullCount(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 full nodes while(len(queue) > 0): node = queue.pop(0) # if it is full node then increment count if node.left is not 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" %(getfullCount(root)) |
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C#
// C# program to count // full 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 full nodes of Tree public class BinaryTree { Node root; /* Function to get the count of full Nodes in a binary tree*/ int getfullCount() { // If tree is empty if (root == null) return 0; // Initialize empty queue. Queue<Node> queue = new Queue<Node>(); // Do level order traversal starting from root queue.Enqueue(root); int count = 0; // Initialize count of full nodes while (queue.Count != 0) { Node temp = queue.Dequeue(); if (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(String []args) { /* 2 / \ 7 5 \ \ 6 9 / \ / 1 11 4 Let us create Binary Tree as shown */ 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.getfullCount()); } } // This code has been contributed by 29AjayKumar |
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Output:
2
Time Complexity: O(n)
Auxiliary Space : O(n) where, n is number of nodes in given binary tree
Method: Recursive
The idea is to traverse the tree in postorder. If the current node is full, we increment result by 1 and add returned values of left and right subtrees.
C++
// C++ program to count full 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 full Nodes in // a binary tree unsigned int getfullCount(struct Node* root) { if (root == NULL) return 0; int res = 0; if (root->left && root->right) res++; res += (getfullCount(root->left) + getfullCount(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 as shown */ 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 << getfullCount(root); return 0; } |
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Java
// Java program to count full 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 full Nodes in // a binary tree static int getfullCount(Node root) { if (root == null) return 0; int res = 0; if (root.left != null && root.right != null) res++; res += (getfullCount(root.left) + getfullCount(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 as shown */ 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(getfullCount(root)); } } |
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Python3
# Python program to count full # nodes in a Binary Tree class newNode(): def __init__(self, data): self.data = data self.left = None self.right = None # Function to get the count of # full Nodes in a binary tree def getfullCount(root): if (root == None): return 0 res = 0 if (root.left and root.right): res += 1 res += (getfullCount(root.left) + getfullCount(root.right)) return res # Driver code if __name__ == '__main__': """ 2 / \ 7 5 \ \ 6 9 / \ / 1 11 4 Let us create Binary Tree as shown """ 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(getfullCount(root)) # This code is contributed by SHUBHAMSINGH10 |
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C#
// C# program to count full 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 full Nodes in // a binary tree static int getfullCount(Node root) { if (root == null) return 0; int res = 0; if (root.left != null && root.right != null) res++; res += (getfullCount(root.left) + getfullCount(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() { /* 2 / \ 7 5 \ \ 6 9 / \ / 1 11 4 Let us create Binary Tree as shown */ 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(getfullCount(root)); } } /* This code contributed by PrinciRaj1992 */ |
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Output:
2
Time Complexity: O(n)
Auxiliary Space : O(n)
where, n is number of nodes in given binary tree
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