Subtree with given sum in a Binary Tree
You are given a binary tree and a given sum. The task is to check if there exist a subtree whose sum of all nodes is equal to the given sum.
Examples :
// For above tree
Input : sum = 17
Output: "Yes"
// sum of all nodes of subtree {3, 5, 9} = 17
Input : sum = 11
Output: "No"
// no subtree with given sum exist
The idea is to traverse tree in Postorder fashion because here we have to think bottom-up . First calculate the sum of left subtree then right subtree and check if sum_left + sum_right + cur_node = sum is satisfying the condition that means any subtree with given sum exist. Below is the recursive implementation of algorithm.
C++
// C++ program to find if there is a subtree with // given sum #include<bits/stdc++.h> using namespace std; /* A binary tree node has data, pointer to left child and a pointer to right child */struct Node { int data; struct Node* left, *right; }; /* utility that allocates a new node with the given data and NULL left and right pointers. */struct Node* newnode(int data) { struct Node* node = new Node; node->data = data; node->left = node->right = NULL; return (node); } // function to check if there exist any subtree with given sum // cur_sum --> sum of current subtree from ptr as root // sum_left --> sum of left subtree from ptr as root // sum_right --> sum of right subtree from ptr as root bool sumSubtreeUtil(struct Node *ptr, int *cur_sum, int sum) { // base condition if (ptr == NULL) { *cur_sum = 0; return false; } // Here first we go to left sub-tree, then right subtree // then first we calculate sum of all nodes of subtree // having ptr as root and assign it as cur_sum // cur_sum = sum_left + sum_right + ptr->data // after that we check if cur_sum == sum int sum_left = 0, sum_right = 0; return ( sumSubtreeUtil(ptr->left, &sum_left, sum) || sumSubtreeUtil(ptr->right, &sum_right, sum) || ((*cur_sum = sum_left + sum_right + ptr->data) == sum)); } // Wrapper over sumSubtreeUtil() bool sumSubtree(struct Node *root, int sum) { // Initialize sum of subtree with root int cur_sum = 0; return sumSubtreeUtil(root, &cur_sum, sum); } // driver program to run the case int main() { struct Node *root = newnode(8); root->left = newnode(5); root->right = newnode(4); root->left->left = newnode(9); root->left->right = newnode(7); root->left->right->left = newnode(1); root->left->right->right = newnode(12); root->left->right->right->right = newnode(2); root->right->right = newnode(11); root->right->right->left = newnode(3); int sum = 22; if (sumSubtree(root, sum)) cout << "Yes"; else cout << "No"; return 0; } |
Java
// Java program to find if there // is a subtree with given sum import java.util.*; class GFG { /* A binary tree node has data, pointer to left child and a pointer to right child */static class Node { int data; Node left, right; } static class INT { int v; INT(int a) { v = a; } } /* utility that allocates a new node with the given data and null left and right pointers. */static Node newnode(int data) { Node node = new Node(); node.data = data; node.left = node.right = null; return (node); } // function to check if there exist // any subtree with given sum // cur_sum -. sum of current subtree // from ptr as root // sum_left -. sum of left subtree // from ptr as root // sum_right -. sum of right subtree // from ptr as root static boolean sumSubtreeUtil(Node ptr, INT cur_sum, int sum) { // base condition if (ptr == null) { cur_sum = new INT(0); return false; } // Here first we go to left // sub-tree, then right subtree // then first we calculate sum // of all nodes of subtree having // ptr as root and assign it as // cur_sum. (cur_sum = sum_left + // sum_right + ptr.data) after that // we check if cur_sum == sum INT sum_left = new INT(0), sum_right = new INT(0); return (sumSubtreeUtil(ptr.left, sum_left, sum) || sumSubtreeUtil(ptr.right, sum_right, sum) || ((cur_sum.v = sum_left.v + sum_right.v + ptr.data) == sum)); } // Wrapper over sumSubtreeUtil() static boolean sumSubtree(Node root, int sum) { // Initialize sum of // subtree with root INT cur_sum = new INT( 0); return sumSubtreeUtil(root, cur_sum, sum); } // Driver Code public static void main(String args[]) { Node root = newnode(8); root.left = newnode(5); root.right = newnode(4); root.left.left = newnode(9); root.left.right = newnode(7); root.left.right.left = newnode(1); root.left.right.right = newnode(12); root.left.right.right.right = newnode(2); root.right.right = newnode(11); root.right.right.left = newnode(3); int sum = 22; if (sumSubtree(root, sum)) System.out.println( "Yes"); else System.out.println( "No"); } } // This code is contributed // by Arnab Kundu |
Python3
# Python3 program to find if there is a # subtree with given sum # Binary Tree Node """ utility that allocates a newNode with the given key """class newnode: # Construct to create a newNode def __init__(self, key): self.data = key self.left = None self.right = None # function to check if there exist any # subtree with given sum # cur_sum -. sum of current subtree # from ptr as root # sum_left -. sum of left subtree from # ptr as root # sum_right -. sum of right subtree # from ptr as root def sumSubtreeUtil(ptr,cur_sum,sum): # base condition if (ptr == None): cur_sum[0] = 0 return False # Here first we go to left sub-tree, # then right subtree then first we # calculate sum of all nodes of subtree # having ptr as root and assign it as cur_sum # cur_sum = sum_left + sum_right + ptr.data # after that we check if cur_sum == sum sum_left, sum_right = [0], [0] x=sumSubtreeUtil(ptr.left, sum_left, sum) y=sumSubtreeUtil(ptr.right, sum_right, sum) cur_sum[0] = (sum_left[0] + sum_right[0] + ptr.data) return ((x or y)or (cur_sum[0] == sum)) # Wrapper over sumSubtreeUtil() def sumSubtree(root, sum): # Initialize sum of subtree with root cur_sum = [0] return sumSubtreeUtil(root, cur_sum, sum) # Driver Code if __name__ == '__main__': root = newnode(8) root.left = newnode(5) root.right = newnode(4) root.left.left = newnode(9) root.left.right = newnode(7) root.left.right.left = newnode(1) root.left.right.right = newnode(12) root.left.right.right.right = newnode(2) root.right.right = newnode(11) root.right.right.left = newnode(3) sum = 22 if (sumSubtree(root, sum)) : print("Yes" ) else: print("No") # This code is contributed by # Shubham Singh(SHUBHAMSINGH10) |
C#
using System; // C# program to find if there // is a subtree with given sum public class GFG { /* 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 class INT { public int v; public INT(int a) { v = a; } } /* utility that allocates a new node with the given data and null left and right pointers. */public static Node newnode(int data) { Node node = new Node(); node.data = data; node.left = node.right = null; return (node); } // function to check if there exist // any subtree with given sum // cur_sum -. sum of current subtree // from ptr as root // sum_left -. sum of left subtree // from ptr as root // sum_right -. sum of right subtree // from ptr as root public static bool sumSubtreeUtil(Node ptr, INT cur_sum, int sum) { // base condition if (ptr == null) { cur_sum = new INT(0); return false; } // Here first we go to left // sub-tree, then right subtree // then first we calculate sum // of all nodes of subtree having // ptr as root and assign it as // cur_sum. (cur_sum = sum_left + // sum_right + ptr.data) after that // we check if cur_sum == sum INT sum_left = new INT(0), sum_right = new INT(0); return (sumSubtreeUtil(ptr.left, sum_left, sum) || sumSubtreeUtil(ptr.right, sum_right, sum) || ((cur_sum.v = sum_left.v + sum_right.v + ptr.data) == sum)); } // Wrapper over sumSubtreeUtil() public static bool sumSubtree(Node root, int sum) { // Initialize sum of // subtree with root INT cur_sum = new INT(0); return sumSubtreeUtil(root, cur_sum, sum); } // Driver Code public static void Main(string[] args) { Node root = newnode(8); root.left = newnode(5); root.right = newnode(4); root.left.left = newnode(9); root.left.right = newnode(7); root.left.right.left = newnode(1); root.left.right.right = newnode(12); root.left.right.right.right = newnode(2); root.right.right = newnode(11); root.right.right.left = newnode(3); int sum = 22; if (sumSubtree(root, sum)) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); } } } // This code is contributed by Shrikant13 |
Output:
Yes
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