# Iterative approach to check for children sum property in a Binary Tree

• Difficulty Level : Medium
• Last Updated : 06 Aug, 2021

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 the sum of data values in left and right children. Consider data value as 0 for NULL children.
Examples:

```Input :
10
/  \
8    2
/ \    \
3   5    2
Output : Yes

Input :
5
/  \
-2    7
/ \    \
1   6    7
Output : No```

We have already discussed the recursive approach. In this post an iterative approach is discussed.
Approach: The idea is to use a queue to do level order traversal of the Binary Tree and simultaneously check for every node:

1. If the current node has two children and if the current node is equal to the sum of its left and right children.
2. If the current node has just left child and if the current node is equal to its left child.
3. If the current node has just right child and if the current node is equal to its right child.

Below is the implementation of the above approach:

## C++

 `// C++ program to check children sum property``#include ``using` `namespace` `std;` `// A binary tree node``struct` `Node {``    ``int` `data;``    ``Node *left, *right;``};` `// Utility function to allocate memory for a new node``Node* newNode(``int` `data)``{``    ``Node* node = ``new` `(Node);``    ``node->data = data;``    ``node->left = node->right = NULL;``    ``return` `(node);``}` `// Function to check if the tree holds``// children sum property``bool` `CheckChildrenSum(Node* root)``{``    ``queue q;` `    ``// Push the root node``    ``q.push(root);` `    ``while` `(!q.empty()) {``        ``Node* temp = q.front();``        ``q.pop();` `        ``// If the current node has both left and right children``        ``if` `(temp->left && temp->right) {``            ``// If the current node is not equal to``            ``// the sum of its left and right children``            ``// return false``            ``if` `(temp->data != temp->left->data + temp->right->data)``                ``return` `false``;` `            ``q.push(temp->left);``            ``q.push(temp->right);``        ``}` `        ``// If the current node has right child``        ``else` `if` `(!temp->left && temp->right) {``            ``// If the current node is not equal to``            ``// its right child return false``            ``if` `(temp->data != temp->right->data)``                ``return` `false``;` `            ``q.push(temp->right);``        ``}` `        ``// If the current node has left child``        ``else` `if` `(!temp->right && temp->left) {``            ``// If the current node is not equal to``            ``// its left child return false``            ``if` `(temp->data != temp->left->data)``                ``return` `false``;` `            ``q.push(temp->left);``        ``}``    ``}` `    ``// If the given tree has children``    ``// sum property return true``    ``return` `true``;``}` `// Driver code``int` `main()``{``    ``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` `(CheckChildrenSum(root))``        ``printf``(``"Yes"``);``    ``else``        ``printf``(``"No"``);` `    ``return` `0;``}`

## Java

 `// Java program to check children sum property``import` `java.util.*;``class` `GFG``{` `// A binary tree node``static` `class` `Node``{``    ``int` `data;``    ``Node left, right;``}` `// Utility function to allocate memory for a new node``static` `Node newNode(``int` `data)``{``    ``Node node = ``new` `Node();``    ``node.data = data;``    ``node.left = node.right = ``null``;``    ``return` `(node);``}` `// Function to check if the tree holds``// children sum property``static` `boolean` `CheckChildrenSum(Node root)``{``    ``Queue q = ``new` `LinkedList();` `    ``// add the root node``    ``q.add(root);` `    ``while` `(q.size() > ``0``)``    ``{``        ``Node temp = q.peek();``        ``q.remove();` `        ``// If the current node has both left and right children``        ``if` `(temp.left != ``null` `&& temp.right != ``null``)``        ``{``            ``// If the current node is not equal to``            ``// the sum of its left and right children``            ``// return false``            ``if` `(temp.data != temp.left.data + temp.right.data)``                ``return` `false``;` `            ``q.add(temp.left);``            ``q.add(temp.right);``        ``}` `        ``// If the current node has right child``        ``else` `if` `(temp.left == ``null` `&& temp.right != ``null``)``        ``{``            ``// If the current node is not equal to``            ``// its right child return false``            ``if` `(temp.data != temp.right.data)``                ``return` `false``;` `            ``q.add(temp.right);``        ``}` `        ``// If the current node has left child``        ``else` `if` `(temp.right == ``null` `&& temp.left != ``null``)``        ``{``            ``// If the current node is not equal to``            ``// its left child return false``            ``if` `(temp.data != temp.left.data)``                ``return` `false``;` `            ``q.add(temp.left);``        ``}``    ``}` `    ``// If the given tree has children``    ``// sum property return true``    ``return` `true``;``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``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` `(CheckChildrenSum(root))``        ``System.out.printf(``"Yes"``);``    ``else``        ``System.out.printf(``"No"``);``}``}` `// This code is contributed by Arnab Kundu`

## Python3

 `# Python3 program to check``# children sum property` `# A binary tree node``class` `Node:``    ` `    ``def` `__init__(``self``, data):``        ``self``.data ``=` `data``        ``self``.left ``=` `None``        ``self``.right ``=` `None` `# Function to check if the tree holds``# children sum property``def` `CheckChildrenSum(root):` `    ``q ``=` `[]``    ` `    ``# Push the root node``    ``q.append(root)` `    ``while` `len``(q) !``=` `0``:``        ``temp ``=` `q.pop()` `        ``# If the current node has both``        ``# left and right children``        ``if` `temp.left ``and` `temp.right:``            ` `            ``# If the current node is not equal``            ``# to the sum of its left and right``            ``# children, return false``            ``if` `(temp.data !``=` `temp.left.data ``+``                             ``temp.right.data):``                ``return` `False` `            ``q.append(temp.left)``            ``q.append(temp.right)``        ` `        ``# If the current node has right child``        ``elif` `not` `temp.left ``and` `temp.right:``            ` `            ``# If the current node is not equal``            ``# to its right child return false``            ``if` `temp.data !``=` `temp.right.data:``                ``return` `False` `            ``q.append(temp.right)``        ` `        ``# If the current node has left child``        ``elif` `not` `temp.right ``and` `temp.left:``            ` `            ``# If the current node is not equal``            ``# to its left child return false``            ``if` `temp.data !``=` `temp.left.data:``                ``return` `False` `            ``q.append(temp.left)` `    ``# If the given tree has children``    ``# sum property return true``    ``return` `True` `# Driver code``if` `__name__ ``=``=` `"__main__"``:` `    ``root ``=` `Node(``10``)``    ``root.left ``=` `Node(``8``)``    ``root.right ``=` `Node(``2``)``    ``root.left.left ``=` `Node(``3``)``    ``root.left.right ``=` `Node(``5``)``    ``root.right.right ``=` `Node(``2``)` `    ``if` `CheckChildrenSum(root):``        ``print``(``"Yes"``)``    ``else``:``        ``print``(``"No"``)` `# This code is contributed``# by Rituraj Jain`

## C#

 `// C# program to check children sum property``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `// A binary tree node``public` `class` `Node``{``    ``public` `int` `data;``    ``public` `Node left, right;``}` `// Utility function to allocate``// memory for a new node``static` `Node newNode(``int` `data)``{``    ``Node node = ``new` `Node();``    ``node.data = data;``    ``node.left = node.right = ``null``;``    ``return` `(node);``}` `// Function to check if the tree holds``// children sum property``static` `Boolean CheckChildrenSum(Node root)``{``    ``Queue q = ``new` `Queue();` `    ``// add the root node``    ``q.Enqueue(root);` `    ``while` `(q.Count > 0)``    ``{``        ``Node temp = q.Peek();``        ``q.Dequeue();` `        ``// If the current node has both``        ``// left and right children``        ``if` `(temp.left != ``null` `&&``            ``temp.right != ``null``)``        ``{``            ``// If the current node is not equal to``            ``// the sum of its left and right children``            ``// return false``            ``if` `(temp.data != temp.left.data +``                             ``temp.right.data)``                ``return` `false``;` `            ``q.Enqueue(temp.left);``            ``q.Enqueue(temp.right);``        ``}` `        ``// If the current node has right child``        ``else` `if` `(temp.left == ``null` `&&``                 ``temp.right != ``null``)``        ``{``            ``// If the current node is not equal to``            ``// its right child return false``            ``if` `(temp.data != temp.right.data)``                ``return` `false``;` `            ``q.Enqueue(temp.right);``        ``}` `        ``// If the current node has left child``        ``else` `if` `(temp.right == ``null` `&&``                 ``temp.left != ``null``)``        ``{``            ``// If the current node is not equal to``            ``// its left child return false``            ``if` `(temp.data != temp.left.data)``                ``return` `false``;` `            ``q.Enqueue(temp.left);``        ``}``    ``}` `    ``// If the given tree has children``    ``// sum property return true``    ``return` `true``;``}` `// Driver code``public` `static` `void` `Main(String []args)``{``    ``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` `(CheckChildrenSum(root))``        ``Console.WriteLine(``"Yes"``);``    ``else``        ``Console.WriteLine(``"No"``);``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output:

`Yes    `

Time Complexity: O(N), where N is the total number of nodes in the binary tree.
Auxiliary Space: O(N)

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