Find Count of Single Valued Subtrees
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.
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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
// 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 |
Python3
# 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) |
C#
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 |
Javascript
<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> |
Count 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; } |
Java
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()); } } |
Python
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()) |
C#
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()); } } |
Javascript
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.
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