Check for Children Sum Property in a Binary Tree
Given a binary tree, write a function that returns true if the tree satisfies below property.
For every node, data value must be equal to sum of data values in left and right children. Consider data value as 0 for NULL children. Below tree is an example
Algorithm:
Traverse the given binary tree. For each node check (recursively) if the node and both its children satisfy the Children Sum Property, if so then return true else return false.
Implementation:
C++
/* Program to check children sum property */#include <bits/stdc++.h>using namespace std; /* A binary tree node has data, left child and right child */struct node{ int data; struct node* left; struct node* right;}; /* returns 1 if children sum property holds for the given node and both of its children*/int isSumProperty(struct node* node){ /* left_data is left child data and right_data is for right child data*/ int left_data = 0, right_data = 0; /* If node is NULL or it's a leaf node then return true */ if(node == NULL || (node->left == NULL && node->right == NULL)) return 1; else { /* If left child is not present then 0 is used as data of left child */ if(node->left != NULL) left_data = node->left->data; /* If right child is not present then 0 is used as data of right child */ if(node->right != NULL) right_data = node->right->data; /* if the node and both of its children satisfy the property return 1 else 0*/ if((node->data == left_data + right_data)&& isSumProperty(node->left) && isSumProperty(node->right)) return 1; else return 0; }} /*Helper function that allocates a new nodewith the given data and NULL left and rightpointers.*/struct node* newNode(int data){ struct node* node = (struct node*)malloc(sizeof(struct node)); node->data = data; node->left = NULL; node->right = NULL; return(node);} // Driver Codeint main(){ struct node *root = newNode(10); root->left = newNode(8); root->right = newNode(2); root->left->left = newNode(3); root->left->right = newNode(5); root->right->right = newNode(2); if(isSumProperty(root)) cout << "The given tree satisfies " << "the children sum property "; else cout << "The given tree does not satisfy " << "the children sum property "; getchar(); return 0;}// This code is contributed by Akanksha Rai |
C
/* Program to check children sum property */#include <stdio.h>#include <stdlib.h> /* A binary tree node has data, left child and right child */struct node{ int data; struct node* left; struct node* right;}; /* returns 1 if children sum property holds for the given node and both of its children*/int isSumProperty(struct node* node){ /* left_data is left child data and right_data is for right child data*/ int left_data = 0, right_data = 0; /* If node is NULL or it's a leaf node then return true */ if(node == NULL || (node->left == NULL && node->right == NULL)) return 1; else { /* If left child is not present then 0 is used as data of left child */ if(node->left != NULL) left_data = node->left->data; /* If right child is not present then 0 is used as data of right child */ if(node->right != NULL) right_data = node->right->data; /* if the node and both of its children satisfy the property return 1 else 0*/ if((node->data == left_data + right_data)&& isSumProperty(node->left) && isSumProperty(node->right)) return 1; else return 0; }} /* 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 = (struct node*)malloc(sizeof(struct node)); node->data = data; node->left = NULL; node->right = NULL; return(node);} /* Driver program to test above function */int main(){ struct node *root = newNode(10); root->left = newNode(8); root->right = newNode(2); root->left->left = newNode(3); root->left->right = newNode(5); root->right->right = newNode(2); if(isSumProperty(root)) printf("The given tree satisfies the children sum property "); else printf("The given tree does not satisfy the children sum property "); getchar(); return 0;} |
Java
// Java program to check children sum property /* A binary tree node has data, pointer to left child and a pointer to right child */class Node { int data; Node left, right; public Node(int d) { data = d; left = right = null; }} class BinaryTree { Node root; /* returns 1 if children sum property holds for the given node and both of its children*/ int isSumProperty(Node node) { /* left_data is left child data and right_data is for right child data*/ int left_data = 0, right_data = 0; /* If node is NULL or it's a leaf node then return true */ if (node == null || (node.left == null && node.right == null)) return 1; else { /* If left child is not present then 0 is used as data of left child */ if (node.left != null) left_data = node.left.data; /* If right child is not present then 0 is used as data of right child */ if (node.right != null) right_data = node.right.data; /* if the node and both of its children satisfy the property return 1 else 0*/ if ((node.data == left_data + right_data) && (isSumProperty(node.left)!=0) && isSumProperty(node.right)!=0) return 1; else return 0; } } /* driver program to test the above functions */ public static void main(String[] args) { BinaryTree tree = new BinaryTree(); tree.root = new Node(10); tree.root.left = new Node(8); tree.root.right = new Node(2); tree.root.left.left = new Node(3); tree.root.left.right = new Node(5); tree.root.right.right = new Node(2); if (tree.isSumProperty(tree.root) != 0) System.out.println("The given tree satisfies children" + " sum property"); else System.out.println("The given tree does not satisfy children" + " sum property"); }} |
Python3
# Python3 program to check children# sum property # Helper class that allocates a new # node with the given data and None # left and right pointers. class newNode: def __init__(self, data): self.data = data self.left = None self.right = None # returns 1 if children sum property # holds for the given node and both # of its childrendef isSumProperty(node): # left_data is left child data and # right_data is for right child data left_data = 0 right_data = 0 # If node is None or it's a leaf # node then return true if(node == None or (node.left == None and node.right == None)): return 1 else: # If left child is not present then # 0 is used as data of left child if(node.left != None): left_data = node.left.data # If right child is not present then # 0 is used as data of right child if(node.right != None): right_data = node.right.data # if the node and both of its children # satisfy the property return 1 else 0 if((node.data == left_data + right_data) and isSumProperty(node.left) and isSumProperty(node.right)): return 1 else: return 0 # Driver Codeif __name__ == '__main__': root = newNode(10) root.left = newNode(8) root.right = newNode(2) root.left.left = newNode(3) root.left.right = newNode(5) root.right.right = newNode(2) if(isSumProperty(root)): print("The given tree satisfies the", "children sum property ") else: print("The given tree does not satisfy", "the children sum property ") # This code is contributed by PranchalK |
C#
// C# program to check children sum property using System; /* 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; public Node(int d) { data = d; left = right = null; }} class GFG{public Node root; /* returns 1 if children sum property holds for the given node and both of its children*/public virtual int isSumProperty(Node node){ /* left_data is left child data and right_data is for right child data*/ int left_data = 0, right_data = 0; /* If node is NULL or it's a leaf node then return true */ if (node == null || (node.left == null && node.right == null)) { return 1; } else { /* If left child is not present then 0 is used as data of left child */ if (node.left != null) { left_data = node.left.data; } /* If right child is not present then 0 is used as data of right child */ if (node.right != null) { right_data = node.right.data; } /* if the node and both of its children satisfy the property return 1 else 0*/ if ((node.data == left_data + right_data) && (isSumProperty(node.left) != 0) && isSumProperty(node.right) != 0) { return 1; } else { return 0; } }} // Driver Codepublic static void Main(string[] args){ GFG tree = new GFG(); tree.root = new Node(10); tree.root.left = new Node(8); tree.root.right = new Node(2); tree.root.left.left = new Node(3); tree.root.left.right = new Node(5); tree.root.right.right = new Node(2); if (tree.isSumProperty(tree.root) != 0) { Console.WriteLine("The given tree satisfies" + " children sum property"); } else { Console.WriteLine("The given tree does not" + " satisfy children sum property"); }}} // This code is contributed by Shrikant13 |
Output:
The given tree satisfies the children sum property
Time Complexity: O(n), we are doing a complete traversal of the tree.
As an exercise, extend the above question for an n-ary tree.
This question was asked by Shekhar.
Please write comments if you find any bug in the above algorithm or a better way to solve the same problem.
