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.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 nodestruct Node{ int data; struct Node* left, *right;};// Utility function to create a new nodeNode* 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 false 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 0int countSingle(Node* root){ // Initialize result int count = 0; // Recursive function to count countSingleRec(root, count); return count;}// Driver program to testint 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 |
Python
# Python program to find the count of single valued subtrees# Node Structureclass 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 0def 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> |
Output:
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.
Thanks to Gaurav Ahirwar for suggesting above solution.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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