Find Count of Single Valued Subtrees
Last Updated :
04 Oct, 2023
Given a binary tree, write a program to count the number of Single Valued Subtrees. A Single Valued Subtree is one in which all the nodes have same value. Expected time complexity is O(n).
Example:
Input: root of below tree
5
/ \
1 5
/ \ \
5 5 5
Output: 4
There are 4 subtrees with single values.
Input: root of below tree
5
/ \
4 5
/ \ \
4 4 5
Output: 5
There are five subtrees with single values.
We strongly recommend you to minimize your browser and try this yourself first.
A Simple Solution is to traverse the tree. For every traversed node, check if all values under this node are same or not. If same, then increment count. Time complexity of this solution is O(n2).
An Efficient Solution is to traverse the tree in bottom up manner. For every subtree visited, return true if subtree rooted under it is single valued and increment count. So the idea is to use count as a reference parameter in recursive calls and use returned values to find out if left and right subtrees are single valued or not.
Below is the implementation of above idea.
C++
// C++ program to find count of single valued subtrees
#include<bits/stdc++.h>
using namespace std;
// A Tree node
struct Node
{
int data;
struct Node* left, *right;
};
// Utility function to create a new node
Node* newNode(int data)
{
Node* temp = new Node;
temp->data = data;
temp->left = temp->right = NULL;
return (temp);
}
// This function increments count by number of single
// valued subtrees under root. It returns true if subtree
// under root is Singly, else false.
bool countSingleRec(Node* root, int &count)
{
// Return true to indicate NULL
if (root == NULL)
return true;
// Recursively count in left and right subtrees also
bool left = countSingleRec(root->left, count);
bool right = countSingleRec(root->right, count);
// If any of the subtrees is not singly, then this
// cannot be singly.
if (left == false || right == false)
return false;
// If left subtree is singly and non-empty, but data
// doesn't match
if (root->left && root->data != root->left->data)
return false;
// Same for right subtree
if (root->right && root->data != root->right->data)
return false;
// If none of the above conditions is true, then
// tree rooted under root is single valued, increment
// count and return true.
count++;
return true;
}
// This function mainly calls countSingleRec()
// after initializing count as 0
int countSingle(Node* root)
{
// Initialize result
int count = 0;
// Recursive function to count
countSingleRec(root, count);
return count;
}
// Driver program to test
int main()
{
/* Let us construct the below tree
5
/ \
4 5
/ \ \
4 4 5 */
Node* root = newNode(5);
root->left = newNode(4);
root->right = newNode(5);
root->left->left = newNode(4);
root->left->right = newNode(4);
root->right->right = newNode(5);
cout << "Count of Single Valued Subtrees is "
<< countSingle(root);
return 0;
}
// Java program to find count of single valued subtrees
/* Class containing left and right child of current
node and key value*/
class Node
{
int data;
Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
class Count
{
int count = 0;
}
class BinaryTree
{
Node root;
Count ct = new Count();
// This function increments count by number of single
// valued subtrees under root. It returns true if subtree
// under root is Singly, else false.
boolean countSingleRec(Node node, Count c)
{
// Return false to indicate NULL
if (node == null)
return true;
// Recursively count in left and right subtrees also
boolean left = countSingleRec(node.left, c);
boolean right = countSingleRec(node.right, c);
// If any of the subtrees is not singly, then this
// cannot be singly.
if (left == false || right == false)
return false;
// If left subtree is singly and non-empty, but data
// doesn't match
if (node.left != null && node.data != node.left.data)
return false;
// Same for right subtree
if (node.right != null && node.data != node.right.data)
return false;
// If none of the above conditions is true, then
// tree rooted under root is single valued, increment
// count and return true.
c.count++;
return true;
}
// This function mainly calls countSingleRec()
// after initializing count as 0
int countSingle()
{
return countSingle(root);
}
int countSingle(Node node)
{
// Recursive function to count
countSingleRec(node, ct);
return ct.count;
}
// Driver program to test above functions
public static void main(String args[])
{
/* Let us construct the below tree
5
/ \
4 5
/ \ \
4 4 5 */
BinaryTree tree = new BinaryTree();
tree.root = new Node(5);
tree.root.left = new Node(4);
tree.root.right = new Node(5);
tree.root.left.left = new Node(4);
tree.root.left.right = new Node(4);
tree.root.right.right = new Node(5);
System.out.println("The count of single valued sub trees is : "
+ tree.countSingle());
}
}
// This code has been contributed by Mayank Jaiswal
# Python program to find the count of single valued subtrees
# Node Structure
class Node:
# Utility function to create a new node
def __init__(self ,data):
self.data = data
self.left = None
self.right = None
# This function increments count by number of single
# valued subtrees under root. It returns true if subtree
# under root is Singly, else false.
def countSingleRec(root , count):
# Return False to indicate None
if root is None :
return True
# Recursively count in left and right subtrees also
left = countSingleRec(root.left , count)
right = countSingleRec(root.right , count)
# If any of the subtrees is not singly, then this
# cannot be singly
if left == False or right == False :
return False
# If left subtree is singly and non-empty , but data
# doesn't match
if root.left and root.data != root.left.data:
return False
# same for right subtree
if root.right and root.data != root.right.data:
return False
# If none of the above conditions is True, then
# tree rooted under root is single valued,increment
# count and return true
count[0] += 1
return True
# This function mainly class countSingleRec()
# after initializing count as 0
def countSingle(root):
# initialize result
count = [0]
# Recursive function to count
countSingleRec(root , count)
return count[0]
# Driver program to test
"""Let us construct the below tree
5
/ \
4 5
/ \ \
4 4 5
"""
root = Node(5)
root.left = Node(4)
root.right = Node(5)
root.left.left = Node(4)
root.left.right = Node(4)
root.right.right = Node(5)
countSingle(root)
print ("Count of Single Valued Subtrees is" , countSingle(root))
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)
using System;
// C# program to find count of single valued subtrees
/* Class containing left and right child of current
node and key value*/
public class Node
{
public int data;
public Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
public class Count
{
public int count = 0;
}
public class BinaryTree
{
public Node root;
public Count ct = new Count();
// This function increments count by number of single
// valued subtrees under root. It returns true if subtree
// under root is Singly, else false.
public virtual bool countSingleRec(Node node, Count c)
{
// Return false to indicate NULL
if (node == null)
{
return true;
}
// Recursively count in left and right subtrees also
bool left = countSingleRec(node.left, c);
bool right = countSingleRec(node.right, c);
// If any of the subtrees is not singly, then this
// cannot be singly.
if (left == false || right == false)
{
return false;
}
// If left subtree is singly and non-empty, but data
// doesn't match
if (node.left != null && node.data != node.left.data)
{
return false;
}
// Same for right subtree
if (node.right != null && node.data != node.right.data)
{
return false;
}
// If none of the above conditions is true, then
// tree rooted under root is single valued, increment
// count and return true.
c.count++;
return true;
}
// This function mainly calls countSingleRec()
// after initializing count as 0
public virtual int countSingle()
{
return countSingle(root);
}
public virtual int countSingle(Node node)
{
// Recursive function to count
countSingleRec(node, ct);
return ct.count;
}
// Driver program to test above functions
public static void Main(string[] args)
{
/* Let us construct the below tree
5
/ \
4 5
/ \ \
4 4 5 */
BinaryTree tree = new BinaryTree();
tree.root = new Node(5);
tree.root.left = new Node(4);
tree.root.right = new Node(5);
tree.root.left.left = new Node(4);
tree.root.left.right = new Node(4);
tree.root.right.right = new Node(5);
Console.WriteLine("The count of single valued sub trees is : " + tree.countSingle());
}
}
// This code is contributed by Shrikant13
<script>
// javascript program to find count of single valued subtrees
/* Class containing left and right child of current
node and key value*/
class Node
{
constructor(item)
{
this.data = item;
this.left = this.right = null;
}
}
class Count
{
constructor(){
this.count = 0;
}
}
var root;
var ct = new Count();
// This function increments count by number of single
// valued subtrees under root. It returns true if subtree
// under root is Singly, else false.
function countSingleRec( node, c)
{
// Return false to indicate NULL
if (node == null)
return true;
// Recursively count in left and right subtrees also
var left = countSingleRec(node.left, c);
var right = countSingleRec(node.right, c);
// If any of the subtrees is not singly, then this
// cannot be singly.
if (left == false || right == false)
return false;
// If left subtree is singly and non-empty, but data
// doesn't match
if (node.left != null && node.data != node.left.data)
return false;
// Same for right subtree
if (node.right != null && node.data != node.right.data)
return false;
// If none of the above conditions is true, then
// tree rooted under root is single valued, increment
// count and return true.
c.count++;
return true;
}
// This function mainly calls countSingleRec()
// after initializing count as 0
function countSingle()
{
return countSingle(root);
}
function countSingle( node)
{
// Recursive function to count
countSingleRec(node, ct);
return ct.count;
}
// Driver program to test above functions
/* Let us construct the below tree
5
/ \
4 5
/ \ \
4 4 5 */
root = new Node(5);
root.left = new Node(4);
root.right = new Node(5);
root.left.left = new Node(4);
root.left.right = new Node(4);
root.right.right = new Node(5);
document.write("The count of single valued sub trees is : "
+ countSingle(root));
// This code contributed by aashish1995
</script>
OutputCount of Single Valued Subtrees is 5
Time complexity of this solution is O(n) where n is number of nodes in given binary tree.
Auxiliary Space: O(h) where h is the height of the tree due to recursion call.
Approach 2: Breadth First Search:
Here's the overall approach of the algorithm using BFS:
- The algorithm includes a helper method to check if a given node is part of a singly valued subtree. It compares the node's value with its left and right child nodes' values.
- The algorithm initializes the count variable to 0.
- It performs a BFS traversal using a queue. It starts by enqueuing the root node.
- Inside the BFS loop, it dequeues a node and checks if it is singly valued.
- If the current node is singly valued, it increments the count variable.
- The algorithm enqueues the left and right child nodes of the current node, if they exist.
- Once the BFS traversal is complete, the count variable contains the total count of single-valued subtrees.
- The algorithm returns the count of single-valued subtrees.
Here is the code of above approach:
C++
#include <iostream>
#include <queue>
using namespace std;
// Class containing left and right child of current
// node and key value
class Node
{
public:
int data;
Node *left, *right;
Node(int item)
{
data = item;
left = right = NULL;
}
};
class Count
{
public:
int count = 0;
};
class BinaryTree
{
public:
Node *root;
Count ct;
// This function increments count by number of single
// valued subtrees under root. It returns true if subtree
// under root is Singly, else false.
bool countSingleRec(Node *node, Count &c)
{
// Return false to indicate NULL
if (node == NULL)
{
return true;
}
// Perform BFS
queue<Node *> q;
q.push(node);
while (!q.empty())
{
Node *curr = q.front();
q.pop();
// Check if the current node is singly valued
if (isSingleValued(curr))
{
c.count++;
}
// Enqueue the left and right child nodes
if (curr->left != NULL)
{
q.push(curr->left);
}
if (curr->right != NULL)
{
q.push(curr->right);
}
}
return true;
}
// Helper function to check if a node is singly valued
bool isSingleValued(Node *node)
{
if (node->left != NULL && node->data != node->left->data)
{
return false;
}
if (node->right != NULL && node->data != node->right->data)
{
return false;
}
return true;
}
// This function mainly calls countSingleRec()
// after initializing count as 0
int countSingle()
{
return countSingle(root);
}
int countSingle(Node *node)
{
// Recursive function to count
countSingleRec(node, ct);
return ct.count;
}
};
// Driver program to test above functions
int main()
{
/* Let us construct the below tree
5
/ \
4 5
/ \ \
4 4 5 */
BinaryTree tree;
tree.root = new Node(5);
tree.root->left = new Node(4);
tree.root->right = new Node(5);
tree.root->left->left = new Node(4);
tree.root->left->right = new Node(4);
tree.root->right->right = new Node(5);
cout << "The count of single valued subtrees is: " << tree.countSingle() << endl;
return 0;
}
import java.util.LinkedList;
import java.util.Queue;
class Node {
int data;
Node left, right;
Node(int item) {
data = item;
left = right = null;
}
}
class Count {
int count = 0;
}
class BinaryTree {
Node root;
Count ct = new Count();
// This function increments count by number of single valued subtrees under root.
// It returns true if subtree under root is singly valued, else false.
boolean countSingleRec(Node node, Count c) {
if (node == null) {
return true;
}
Queue<Node> queue = new LinkedList<>();
queue.add(node);
while (!queue.isEmpty()) {
Node curr = queue.poll();
if (isSingleValued(curr)) {
c.count++;
}
if (curr.left != null) {
queue.add(curr.left);
}
if (curr.right != null) {
queue.add(curr.right);
}
}
return true;
}
// Helper function to check if a node is singly valued
boolean isSingleValued(Node node) {
if (node.left != null && node.data != node.left.data) {
return false;
}
if (node.right != null && node.data != node.right.data) {
return false;
}
return true;
}
// This function mainly calls countSingleRec() after initializing count as 0
int countSingle() {
return countSingleRec(root, ct) ? ct.count : 0;
}
// Driver program to test above functions
public static void main(String[] args) {
// Let us construct the below tree
// 5
// / \
// 4 5
// / \ \
// 4 4 5
BinaryTree tree = new BinaryTree();
tree.root = new Node(5);
tree.root.left = new Node(4);
tree.root.right = new Node(5);
tree.root.left.left = new Node(4);
tree.root.left.right = new Node(4);
tree.root.right.right = new Node(5);
System.out.println("The count of single valued subtrees is: " + tree.countSingle());
}
}
from collections import deque
class Node:
def __init__(self, item):
self.data = item
self.left = None
self.right = None
class Count:
def __init__(self):
self.count = 0
class BinaryTree:
def __init__(self):
self.root = None
self.ct = Count()
def countSingleRec(self, node, c):
if node is None:
return True
q = deque()
q.append(node)
while q:
curr = q.popleft()
if self.isSingleValued(curr):
c.count += 1
if curr.left:
q.append(curr.left)
if curr.right:
q.append(curr.right)
return True
def isSingleValued(self, node):
if node.left and node.data != node.left.data:
return False
if node.right and node.data != node.right.data:
return False
return True
def countSingle(self):
self.countSingleRec(self.root, self.ct)
return self.ct.count
if __name__ == "__main__":
# Construct the tree
tree = BinaryTree()
tree.root = Node(5)
tree.root.left = Node(4)
tree.root.right = Node(5)
tree.root.left.left = Node(4)
tree.root.left.right = Node(4)
tree.root.right.right = Node(5)
print("The count of single valued subtrees is:", tree.countSingle())
using System;
using System.Collections.Generic;
/* Class containing left and right child of current
node and key value*/
public class Node
{
public int data;
public Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
public class Count
{
public int count = 0;
}
public class BinaryTree
{
public Node root;
public Count ct = new Count();
// This function increments count by number of single
// valued subtrees under root. It returns true if subtree
// under root is Singly, else false.
public virtual bool countSingleRec(Node node, Count c)
{
// Return false to indicate NULL
if (node == null)
{
return true;
}
// Perform BFS
Queue<Node> queue = new Queue<Node>();
queue.Enqueue(node);
while (queue.Count > 0)
{
Node curr = queue.Dequeue();
// Check if the current node is singly valued
if (isSingleValued(curr))
{
c.count++;
}
// Enqueue the left and right child nodes
if (curr.left != null)
{
queue.Enqueue(curr.left);
}
if (curr.right != null)
{
queue.Enqueue(curr.right);
}
}
return true;
}
// Helper function to check if a node is singly valued
private bool isSingleValued(Node node)
{
if (node.left != null && node.data != node.left.data)
{
return false;
}
if (node.right != null && node.data != node.right.data)
{
return false;
}
return true;
}
// This function mainly calls countSingleRec()
// after initializing count as 0
public virtual int countSingle()
{
return countSingle(root);
}
public virtual int countSingle(Node node)
{
// Recursive function to count
countSingleRec(node, ct);
return ct.count;
}
// Driver program to test above functions
public static void Main(string[] args)
{
/* Let us construct the below tree
5
/ \
4 5
/ \ \
4 4 5 */
BinaryTree tree = new BinaryTree();
tree.root = new Node(5);
tree.root.left = new Node(4);
tree.root.right = new Node(5);
tree.root.left.left = new Node(4);
tree.root.left.right = new Node(4);
tree.root.right.right = new Node(5);
Console.WriteLine("The count of single valued subtrees is: " + tree.countSingle());
}
}
class Node {
constructor(item) {
this.data = item;
this.left = this.right = null;
}
}
class Count {
constructor() {
this.count = 0;
}
}
class BinaryTree {
constructor() {
this.root = null;
this.ct = new Count();
}
// This function increments count by the number of single
// valued subtrees under root. It returns true if the subtree
// under root is singly, else false.
countSingleRec(node, c) {
// Return true to indicate null node
if (node === null) {
return true;
}
// Perform BFS
const queue = [];
queue.push(node);
while (queue.length !== 0) {
const curr = queue.shift();
// Check if the current node is singly valued
if (this.isSingleValued(curr)) {
c.count++;
}
// Enqueue the left and right child nodes
if (curr.left !== null) {
queue.push(curr.left);
}
if (curr.right !== null) {
queue.push(curr.right);
}
}
return true;
}
// Helper function to check if a node is singly valued
isSingleValued(node) {
if (node.left !== null && node.data !== node.left.data) {
return false;
}
if (node.right !== null && node.data !== node.right.data) {
return false;
}
return true;
}
// This function mainly calls countSingleRec()
// after initializing count as 0
countSingle() {
return this.countSingle(this.root);
}
countSingle(node) {
// Recursive function to count
this.countSingleRec(node, this.ct);
return this.ct.count;
}
}
// Driver program to test above functions
function main() {
/* Let us construct the below tree
5
/ \
4 5
/ \ \
4 4 5 */
const tree = new BinaryTree();
tree.root = new Node(5);
tree.root.left = new Node(4);
tree.root.right = new Node(5);
tree.root.left.left = new Node(4);
tree.root.left.right = new Node(4);
tree.root.right.right = new Node(5);
console.log("The count of single valued subtrees is:", tree.countSingle());
}
main();
Output
The count of single valued sub trees is : 5
Time complexity: O(n), where n is the number of nodes in given binary tree.
Auxiliary Space: O(h), where h is the height of the tree due to recursion call.
Similar Reads
Count of subtrees possible from an N-ary Tree
Given an N-ary tree consisting of N nodes having values 0 to (N - 1), the task is to find the total number of subtrees present in the given tree. Since the count can be very large, so print the count modulo 1000000007. Examples: Input: N = 3 0 / 1 /2Output: 7Explanation:The total number of subtrees
9 min read
Count BST subtrees that lie in given range
Given a Binary Search Tree (BST) and a range [low, high], the task is to count the number of nodes where all the nodes under that node (or subtree rooted with that node) lie in the given range. Examples: Input : low = 8, high = 10 Output: 1 Explanation: There is only 1 node with the value 10 whose s
7 min read
Find Maximum Subtree of the Same Color
Given a 2D integer array edges[] representing a tree with n nodes, numbered from 0 to n - 1, rooted at node 0, where edges[i] = [ui, vi] means there is an edge between the nodes vi and ui. Another array colors[] of size n is also given, where colors[i] is the color assigned to node i. We have to fin
15 min read
Find largest subtree sum in a tree
Given a Binary Tree, the task is to find a subtree with the maximum sum in the tree.Examples: Input: Output: 28Explanation: As all the tree elements are positive, the largest subtree sum is equal to sum of all tree elements.Input: Output: 7Explanation: Subtree with largest sum is: Table of Content[E
15+ min read
Count of subtrees in a Binary Tree having XOR value K
Given a value K and a binary tree, the task is to find out number of subtrees having XOR of all its elements equal to K.Examples: Input K = 5, Tree = 2 / \ 1 9 / \ 10 5Output: 2Explanation:Subtree 1: 5It has only one element i.e. 5.So XOR of subtree = 5Subtree 1: 2 / \ 1 9 / \ 10 5It has elements 2,
14 min read
Count Number of Nodes With Value One in Undirected Tree
Given an undirected connected tree with n nodes labeled from 1 to n, and an integer array of queries[]. Initially, all nodes have a value of 0. For each query in the array, you need to flip the values of all nodes in the subtree of the node with the corresponding label. The parent of a node with lab
8 min read
Even size subtree in n-ary tree
Given an n-ary tree of n vertices and n-1 edges. The tree is given in the form of adjacency list. Find number of subtrees of even size in given tree. Examples: Input : 1 / \ 2 3 / \ \ 4 5 6 / \ 7 8 Output : 2 Subtree rooted at 1 and 3 are of even size. Input : 1 / \ 2 3 / | \ \ 4 5 6 7 / | \ 8 9 10
7 min read
Find the type of Tree based on the child count
Given a tree that has some nodes and each node has a certain number of child nodes, the task is to find the type of tree based on the child count. For example, if each node has two children, then we can say it is a binary tree, and if each node has almost three children, then we can say it is a Tern
7 min read
Count of subtrees in a Binary Tree having bitwise OR value K
Given a value K and a binary tree, the task is to find out the number of subtrees having bitwise OR of all its elements equal to K. Examples: Input: K = 5, Tree = 2 / \ 1 1 / \ \ 10 5 4 Output: 2 Explanation: Subtree 1: 5 It has only one element i.e. 5. So bitwise OR of subtree = 5 Subtree 2: 1 \ 4
8 min read
Remove all subtrees consisting only of even valued nodes from a Binary Tree
Given a Binary Tree, the task is to remove all the subtrees that do not contain any odd valued node. Print the Levelorder Traversal of the tree after removal of these subtrees. Note: Print NULL for removed nodes. Examples: Input: Below is the given Tree: 1 \ 2 / \ 8 5Output: 1 NULL 2 NULL 5Explanati
8 min read