Count the unique type of nodes present in Binary tree

Last Updated : 23 Jul, 2025

According to the property of a Binary tree, a node can have at most two children so there are three cases where a node can have two children, one child, or no child, the task is to track the count of unique nodes and return the total number of unique nodes that have no child, one child, and two children. Given the root of the binary tree return a vector where arr[0] represents the total unique nodes that contain 0 children, arr[1] represents the total unique nodes having 1 child, and arr[2] represents the total unique nodes that have exactly two children.

Examples:

Input:

2
/ \
1 4
/ \ \
2 1 1
/ \
7 5

Output: 4 1 2
Explanation: Nodes with no child are 2, 1, 7, and 5 and all these nodes have unique values so the count is 4. There is only 1 Node with 1 child i.e. 4 so the count will be 1 here.
Nodes with two children are 2 (root), 1 ( 2's child ), and 1 ( 4's child ), since 1 is repeating that's why only 2 (root) and 1 will be counted as unique nodes so the count will be 2 here.

Input:

3
/ \
2 2

Output: 1 0 1
Explanation: Nodes with no child are 2 and 2 so the unique node's count will be 1.
There is no node with 1 child so the count will be 0.
There is only 1 node with 2 children i.e. 3 (root node) so the count will be 1 here.

Approach: To solve the problem follow the below idea:

We can consider the different types of nodes as a pattern and use a hashmap to store every unique node's pattern.

Follow the steps to solve the problem:

We can use three hashmaps for three patterns i.e. node with two children, node with one child, node with no child, and traverse over the tree and match the current node's pattern if the current node's value is unique then store it in the hashmap.
After completing the entire traversal return the size of all three hashmaps which represent all three patterns. For storing data in hashmaps pass the hashmap by reference.

Below code is the implementation of the above approach:

C++

// C++ program to count the unique
// nodes in Binary tree
#include <bits/stdc++.h>
using namespace std;

// A binary tree node has data,
// left child and right child
struct node {
    int data;
    node* left;
    node* right;
};

// Helper function for inserting values in
// hashmap and tracking pattern
void trackPattern(node* root, map<int, int>& twoChild,
                map<int, int>& oneChild,
                map<int, int>& noChild)
{

    // When node have two child
    if (root->left && root->right) {

        // Insert the node into
        // hashmap
        twoChild[root->data]++;

        // Call recursion on left child
        trackPattern(root->left, twoChild, oneChild,
                    noChild);

        // Call recursion on right child
        trackPattern(root->right, twoChild, oneChild,
                    noChild);
    }

    // When node is leaf node
    else if (!root->left && !root->right) {

        // Insert the node into
        // hashmap
        noChild[root->data]++;
    }

    // When node have one child
    else {
        oneChild[root->data]++;

        // If right child is null then
        // call recursion on left child
        if (root->left)
            trackPattern(root->left, twoChild, oneChild,
                        noChild);

        // If left child is null then call
        // recursion on right child
        if (root->right)
            trackPattern(root->right, twoChild, oneChild,
                        noChild);
    }
}

// Main function for counting total unique nodes
vector<int> findUniquePattern(node* root)
{

    // Initializing hashmap for tracking
    // different patterns
    map<int, int> twoChild, oneChild, noChild;

    // Function call for tracking the
    // unique nodes
    trackPattern(root, twoChild, oneChild, noChild);

    // Returning the size of hashmap which total
    // represents unique nodes
    return { (int)noChild.size(), (int)oneChild.size(),
            (int)twoChild.size() };
}

// Helper function that allocates a new node
// with the given data and NULL left
// and right pointers.
node* newNode(int data)
{
    node* node1 = new node();
    node1->data = data;
    node1->left = NULL;
    node1->right = NULL;
    return (node1);
}

// Driver code
int main()
{
    node* root = newNode(2);
    root->left = newNode(1);
    root->right = newNode(4);
    root->left->left = newNode(2);
    root->left->right = newNode(1);
    root->right->right = newNode(1);
    root->right->right->left = newNode(7);
    root->right->right->right = newNode(5);

    // Function call
    vector<int> uniqueNodes = findUniquePattern(root);

    // Printing the values of unique
    // nodes
    cout << uniqueNodes[0] << " " << uniqueNodes[1] << " "
        << uniqueNodes[2];
    return 0;
}

Java

import java.util.*;

class Node {
    int data;
    Node left;
    Node right;

    Node(int data) {
        this.data = data;
        left = null;
        right = null;
    }
}

public class Main {
    public static void main(String[] args) {
        Node root = new Node(2);
        root.left = new Node(1);
        root.right = new Node(4);
        root.left.left = new Node(2);
        root.left.right = new Node(1);
        root.right.right = new Node(1);
        root.right.right.left = new Node(7);
        root.right.right.right = new Node(5);

        // Function call
        List<Integer> uniqueNodes = findUniquePattern(root);

        // Printing the values of unique nodes
        System.out.println(uniqueNodes.get(0) + " " +
                           uniqueNodes.get(1) + " " +
                           uniqueNodes.get(2));
    }

    // Function for counting total unique nodes
    public static List<Integer> findUniquePattern(Node root) {
        // Initializing hashmaps for tracking different patterns
        Map<Integer, Integer> twoChild = new HashMap<>();
        Map<Integer, Integer> oneChild = new HashMap<>();
        Map<Integer, Integer> noChild = new HashMap<>();

        // Function call for tracking the unique nodes
        trackPattern(root, twoChild, oneChild, noChild);

        // Returning the size of hashmaps, which represents unique nodes
        return Arrays.asList(noChild.size(), oneChild.size(), twoChild.size());
    }

    // Helper function for inserting values in hashmaps and tracking pattern
    public static void trackPattern(Node root, Map<Integer, Integer> twoChild,
                                    Map<Integer, Integer> oneChild,
                                    Map<Integer, Integer> noChild) {
        if (root == null) {
            return;
        }

        // When node have two child
        if (root.left != null && root.right != null) {
            // Insert the node into hashmap
            twoChild.put(root.data, twoChild.getOrDefault(root.data, 0) + 1);

            // Call recursion on left child
            trackPattern(root.left, twoChild, oneChild, noChild);

            // Call recursion on right child
            trackPattern(root.right, twoChild, oneChild, noChild);
        }

        // When node is leaf node
        else if (root.left == null && root.right == null) {
            // Insert the node into hashmap
            noChild.put(root.data, noChild.getOrDefault(root.data, 0) + 1);
        }

        // When node have one child
        else {
            oneChild.put(root.data, oneChild.getOrDefault(root.data, 0) + 1);

            // If right child is not null, call recursion on right child
            if (root.left != null) {
                trackPattern(root.left, twoChild, oneChild, noChild);
            }

            // If left child is not null, call recursion on left child
            if (root.right != null) {
                trackPattern(root.right, twoChild, oneChild, noChild);
            }
        }
    }
}

Python3

# Node class
class Node:
    # Constructor to initialize data
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None

# Function for counting total unique nodes
def findUniquePattern(root):
    # Initializing hashmaps for tracking different patterns
    twoChild = {}
    oneChild = {}
    noChild = {}

    # Function call for tracking the unique nodes
    trackPattern(root, twoChild, oneChild, noChild)

    # Returning the size of hashmaps, which represents unique nodes
    return [len(noChild), len(oneChild), len(twoChild)]

# Helper function for inserting values in hashmaps and tracking pattern
def trackPattern(root, twoChild, oneChild, noChild):
    if root is None:
        return

    # When node have two child
    if root.left is not None and root.right is not None:
        # Insert the node into hashmap
        twoChild[root.data] = twoChild.get(root.data, 0) + 1

        # Call recursion on left child
        trackPattern(root.left, twoChild, oneChild, noChild)

        # Call recursion on right child
        trackPattern(root.right, twoChild, oneChild, noChild)

    # When node is leaf node
    elif root.left is None and root.right is None:
        # Insert the node into hashmap
        noChild[root.data] = noChild.get(root.data, 0) + 1

    # When node have one child
    else:
        oneChild[root.data] = oneChild.get(root.data, 0) + 1

        # If right child is not None, call recursion on right child
        if root.left is not None:
            trackPattern(root.left, twoChild, oneChild, noChild)

        # If left child is not None, call recursion on left child
        if root.right is not None:
            trackPattern(root.right, twoChild, oneChild, noChild)


# Driver code
if __name__ == '__main__':
    root = Node(2)
    root.left = Node(1)
    root.right = Node(4)
    root.left.left = Node(2)
    root.left.right = Node(1)
    root.right.right = Node(1)
    root.right.right.left = Node(7)
    root.right.right.right = Node(5)

    # Function call
    uniqueNodes = findUniquePattern(root)

    # Printing the values of unique nodes
    print(uniqueNodes[0], uniqueNodes[1], uniqueNodes[2])

using System;
using System.Collections.Generic;

class Node
{
    public int Data;
    public Node Left, Right;
    // Constructor to create a new node
    public Node(int item)
    {
        Data = item;
        Left = Right = null;
    }
}
class GFG
{
    // Helper function for the inserting values in the hashmap and tracking patterns
    private static void TrackPattern(Node root, Dictionary<int, int> twoChild,
                                    Dictionary<int, int> oneChild,
                                    Dictionary<int, int> noChild)
    {
        if (root == null)
            return;
              // When the node has two children
        if (root.Left != null && root.Right != null)
        {
            twoChild[root.Data] = twoChild.ContainsKey(root.Data) ? twoChild[root.Data] + 1 : 1;
            TrackPattern(root.Left, twoChild, oneChild, noChild);
            TrackPattern(root.Right, twoChild, oneChild, noChild);
        }
                 // When the node is a leaf node
        else if (root.Left == null && root.Right == null)
        {
            noChild[root.Data] = noChild.ContainsKey(root.Data) ? noChild[root.Data] + 1 : 1;
        }
              // When the node has one child
        else
        {
                      // Insert the node into the hashmap
            oneChild[root.Data] = oneChild.ContainsKey(root.Data) ? oneChild[root.Data] + 1 : 1;
            if (root.Left != null)
                TrackPattern(root.Left, twoChild, oneChild, noChild);
            if (root.Right != null)
                TrackPattern(root.Right, twoChild, oneChild, noChild);
        }
    }
    private static List<int> FindUniquePattern(Node root)
    {
         // Initializing hashmap for tracking
        // different patterns
        Dictionary<int, int> twoChild = new Dictionary<int, int>();
        Dictionary<int, int> oneChild = new Dictionary<int, int>();
        Dictionary<int, int> noChild = new Dictionary<int, int>();
        TrackPattern(root, twoChild, oneChild, noChild);
        return new List<int> { noChild.Count, oneChild.Count, twoChild.Count };
    }
    public static void Main(string[] args)
    {
          // Driver code
        Node root = new Node(2);
        root.Left = new Node(1);
        root.Right = new Node(4);
        root.Left.Left = new Node(2);
        root.Left.Right = new Node(1);
        root.Right.Right = new Node(1);
        root.Right.Right.Left = new Node(7);
        root.Right.Right.Right = new Node(5);
        List<int> uniqueNodes = FindUniquePattern(root);
              // Printing the values of unique nodes
        Console.WriteLine($"{uniqueNodes[0]} {uniqueNodes[1]} {uniqueNodes[2]}");
    }
}

JavaScript

// Define a binary tree node structure
class TreeNode {
    constructor(data) {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}

// Function to count unique nodes in a binary tree
function findUniquePattern(root) {
    const noChild = new Map();  // To store nodes with no children
    const oneChild = new Map(); // To store nodes with one child
    const twoChild = new Map(); // To store nodes with two children

    // Helper function for inserting values in maps and tracking pattern
    function trackPattern(node) {
        if (!node) {
            return;
        }

        // When the node has two children
        if (node.left && node.right) {
            // Insert the node into the twoChild map
            if (!twoChild.has(node.data)) {
                twoChild.set(node.data, 0);
            }
            twoChild.set(node.data, twoChild.get(node.data) + 1);

            // Recursively call on left child
            trackPattern(node.left);

            // Recursively call on right child
            trackPattern(node.right);
        }
        // When the node is a leaf node
        else if (!node.left && !node.right) {
            // Insert the node into the noChild map
            if (!noChild.has(node.data)) {
                noChild.set(node.data, 0);
            }
            noChild.set(node.data, noChild.get(node.data) + 1);
        }
        // When the node has one child
        else {
            // Insert the node into the oneChild map
            if (!oneChild.has(node.data)) {
                oneChild.set(node.data, 0);
            }
            oneChild.set(node.data, oneChild.get(node.data) + 1);

            // If the right child is not null, call recursively on it
            if (node.left) {
                trackPattern(node.left);
            }

            // If the left child is not null, call recursively on it
            if (node.right) {
                trackPattern(node.right);
            }
        }
    }

    // Start tracking unique nodes from the root
    trackPattern(root);

    // Return the counts of unique nodes in an array
    return [noChild.size, oneChild.size, twoChild.size];
}

// Create the binary tree
const root = new TreeNode(2);
root.left = new TreeNode(1);
root.right = new TreeNode(4);
root.left.left = new TreeNode(2);
root.left.right = new TreeNode(1);
root.right.right = new TreeNode(1);
root.right.right.left = new TreeNode(7);
root.right.right.right = new TreeNode(5);

// Function call
const uniqueNodes = findUniquePattern(root);

// Print the values of unique nodes
console.log(uniqueNodes[0] + " " + uniqueNodes[1] + " " + uniqueNodes[2]);

Output

4 1 2

Time Complexity: O(N*logN), N For traversing the tree and LogN for map operations
Auxiliary Space: O(N), For hashmaps, at max, N values can be stored

Analysis of Algorithms

nichil942singh

Improve

Article Tags :

Practice Tags :

Count the unique type of nodes present in Binary tree

Similar Reads

Basics & Prerequisites

Data Structures

Algorithms

Advanced

Interview Preparation

Practice Problem

Thank You!

What kind of Experience do you want to share?