# Find sum of all left leaves in a given Binary Tree

Given a Binary Tree, find sum of all left leaves in it. For example, sum of all left leaves in below Binary Tree is 5+1=6. ## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

The idea is to traverse the tree, starting from root. For every node, check if its left subtree is a leaf. If it is, then add it to the result.

Following is the implementation of above idea.

## C++

 `// A C++ program to find sum of all left leaves ` `#include ` `using` `namespace` `std; ` ` `  `/* A binary tree Node has key, pointer to left and right ` `   ``children */` `struct` `Node ` `{ ` `    ``int` `key; ` `    ``struct` `Node* left, *right; ` `}; ` ` `  `/* Helper function that allocates a new node with the ` `   ``given data and NULL left and right pointer. */` `Node *newNode(``char` `k) ` `{ ` `    ``Node *node = ``new` `Node; ` `    ``node->key = k; ` `    ``node->right = node->left = NULL; ` `    ``return` `node; ` `} ` ` `  `// A utility function to check if a given node is leaf or not ` `bool` `isLeaf(Node *node) ` `{ ` `   ``if` `(node == NULL) ` `       ``return` `false``; ` `   ``if` `(node->left == NULL && node->right == NULL) ` `       ``return` `true``; ` `   ``return` `false``; ` `} ` ` `  `// This function returns sum of all left leaves in a given ` `// binary tree ` `int` `leftLeavesSum(Node *root) ` `{ ` `    ``// Initialize result ` `    ``int` `res = 0; ` ` `  `    ``// Update result if root is not NULL ` `    ``if` `(root != NULL) ` `    ``{ ` `       ``// If left of root is NULL, then add key of ` `       ``// left child ` `       ``if` `(isLeaf(root->left)) ` `            ``res += root->left->key; ` `       ``else` `// Else recur for left child of root ` `            ``res += leftLeavesSum(root->left); ` ` `  `       ``// Recur for right child of root and update res ` `       ``res += leftLeavesSum(root->right); ` `    ``} ` ` `  `    ``// return result ` `    ``return` `res; ` `} ` ` `  `/* Driver program to test above functions*/` `int` `main() ` `{ ` `    ``// Let us a construct the Binary Tree ` `    ``struct` `Node *root         = newNode(20); ` `    ``root->left                = newNode(9); ` `    ``root->right               = newNode(49); ` `    ``root->right->left         = newNode(23); ` `    ``root->right->right        = newNode(52); ` `    ``root->right->right->left  = newNode(50); ` `    ``root->left->left          = newNode(5); ` `    ``root->left->right         = newNode(12); ` `    ``root->left->right->right  = newNode(12); ` `    ``cout << ``"Sum of left leaves is "` `         ``<< leftLeavesSum(root); ` `    ``return` `0; ` `}`

## Java

 `// Java program to find sum of all left leaves ` `class` `Node  ` `{ ` `    ``int` `data; ` `    ``Node left, right; ` `  `  `    ``Node(``int` `item)  ` `    ``{ ` `        ``data = item; ` `        ``left = right = ``null``; ` `    ``} ` `} ` `  `  `class` `BinaryTree  ` `{ ` `    ``Node root; ` `  `  `    ``// A utility function to check if a given node is leaf or not ` `    ``boolean` `isLeaf(Node node)  ` `    ``{ ` `        ``if` `(node == ``null``) ` `            ``return` `false``; ` `        ``if` `(node.left == ``null` `&& node.right == ``null``) ` `            ``return` `true``; ` `        ``return` `false``; ` `    ``} ` `  `  `     ``// This function returns sum of all left leaves in a given ` `     ``// binary tree ` `    ``int` `leftLeavesSum(Node node)  ` `    ``{ ` `        ``// Initialize result ` `        ``int` `res = ``0``; ` `  `  `        ``// Update result if root is not NULL ` `        ``if` `(node != ``null``)  ` `        ``{ ` `            ``// If left of root is NULL, then add key of ` `            ``// left child ` `            ``if` `(isLeaf(node.left)) ` `                ``res += node.left.data; ` `            ``else` `// Else recur for left child of root ` `                ``res += leftLeavesSum(node.left); ` `  `  `            ``// Recur for right child of root and update res ` `            ``res += leftLeavesSum(node.right); ` `        ``} ` `  `  `        ``// return result ` `        ``return` `res; ` `    ``} ` `  `  `    ``// Driver program ` `    ``public` `static` `void` `main(String args[])  ` `    ``{ ` `        ``BinaryTree tree = ``new` `BinaryTree(); ` `        ``tree.root = ``new` `Node(``20``); ` `        ``tree.root.left = ``new` `Node(``9``); ` `        ``tree.root.right = ``new` `Node(``49``); ` `        ``tree.root.left.right = ``new` `Node(``12``); ` `        ``tree.root.left.left = ``new` `Node(``5``); ` `        ``tree.root.right.left = ``new` `Node(``23``); ` `        ``tree.root.right.right = ``new` `Node(``52``); ` `        ``tree.root.left.right.right = ``new` `Node(``12``); ` `        ``tree.root.right.right.left = ``new` `Node(``50``); ` `  `  `        ``System.out.println(``"The sum of leaves is "` `+  ` `                                       ``tree.leftLeavesSum(tree.root)); ` `    ``} ` `} ` `  `  `// This code is contributed by Mayank Jaiswal  `

## Python

 `# Python program to find sum of all left leaves ` ` `  `# A Binary tree node ` `class` `Node: ` `    ``# Constructor to create a new Node ` `    ``def` `__init__(``self``, key): ` `        ``self``.key ``=` `key ` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None` ` `  `# A utility function to check if a given node is leaf or not ` `def` `isLeaf(node): ` `    ``if` `node ``is` `None``: ` `        ``return` `False` `    ``if` `node.left ``is` `None` `and` `node.right ``is` `None``: ` `        ``return` `True` `    ``return` `False` ` `  `# This function return sum of all left leaves in a ` `# given binary tree ` `def` `leftLeavesSum(root): ` ` `  `    ``# Initialize result ` `    ``res ``=` `0` `     `  `    ``# Update result if root is not None ` `    ``if` `root ``is` `not` `None``: ` ` `  `        ``# If left of root is None, then add key of ` `        ``# left child ` `        ``if` `isLeaf(root.left): ` `            ``res ``+``=` `root.left.key ` `        ``else``: ` `            ``# Else recur for left child of root ` `            ``res ``+``=` `leftLeavesSum(root.left) ` ` `  `        ``# Recur for right child of root and update res ` `        ``res ``+``=` `leftLeavesSum(root.right) ` `    ``return` `res ` ` `  `# Driver program to test above function  ` ` `  `# Let us constrcut the Binary Tree shown in the above function ` `root ``=` `Node(``20``) ` `root.left ``=` `Node(``9``) ` `root.right ``=` `Node(``49``) ` `root.right.left ``=` `Node(``23``)         ` `root.right.right ``=` `Node(``52``) ` `root.right.right.left ``=` `Node(``50``) ` `root.left.left ``=` `Node(``5``) ` `root.left.right ``=` `Node(``12``) ` `root.left.right.right ``=` `Node(``12``) ` `print` `"Sum of left leaves is"``, leftLeavesSum(root) ` ` `  `# This code is contributed by Nikhil Kumar Singh(nickzuck_007) `

## C#

 `using` `System; ` ` `  `// C# program to find sum of all left leaves  ` `public` `class` `Node ` `{ ` `    ``public` `int` `data; ` `    ``public` `Node left, right; ` ` `  `    ``public` `Node(``int` `item) ` `    ``{ ` `        ``data = item; ` `        ``left = right = ``null``; ` `    ``} ` `} ` ` `  `public` `class` `BinaryTree ` `{ ` `    ``public` `Node root; ` ` `  `    ``// A utility function to check if a given node is leaf or not  ` `    ``public` `virtual` `bool` `isLeaf(Node node) ` `    ``{ ` `        ``if` `(node == ``null``) ` `        ``{ ` `            ``return` `false``; ` `        ``} ` `        ``if` `(node.left == ``null` `&& node.right == ``null``) ` `        ``{ ` `            ``return` `true``; ` `        ``} ` `        ``return` `false``; ` `    ``} ` ` `  `     ``// This function returns sum of all left leaves in a given  ` `     ``// binary tree  ` `    ``public` `virtual` `int` `leftLeavesSum(Node node) ` `    ``{ ` `        ``// Initialize result  ` `        ``int` `res = 0; ` ` `  `        ``// Update result if root is not NULL  ` `        ``if` `(node != ``null``) ` `        ``{ ` `            ``// If left of root is NULL, then add key of  ` `            ``// left child  ` `            ``if` `(isLeaf(node.left)) ` `            ``{ ` `                ``res += node.left.data; ` `            ``} ` `            ``else` `// Else recur for left child of root ` `            ``{ ` `                ``res += leftLeavesSum(node.left); ` `            ``} ` ` `  `            ``// Recur for right child of root and update res  ` `            ``res += leftLeavesSum(node.right); ` `        ``} ` ` `  `        ``// return result  ` `        ``return` `res; ` `    ``} ` ` `  `    ``// Driver program  ` `    ``public` `static` `void` `Main(``string``[] args) ` `    ``{ ` `        ``BinaryTree tree = ``new` `BinaryTree(); ` `        ``tree.root = ``new` `Node(20); ` `        ``tree.root.left = ``new` `Node(9); ` `        ``tree.root.right = ``new` `Node(49); ` `        ``tree.root.left.right = ``new` `Node(12); ` `        ``tree.root.left.left = ``new` `Node(5); ` `        ``tree.root.right.left = ``new` `Node(23); ` `        ``tree.root.right.right = ``new` `Node(52); ` `        ``tree.root.left.right.right = ``new` `Node(12); ` `        ``tree.root.right.right.left = ``new` `Node(50); ` ` `  `        ``Console.WriteLine(``"The sum of leaves is "` `+ tree.leftLeavesSum(tree.root)); ` `    ``} ` `} ` ` `  `  ``//  This code is contributed by Shrikant13 `

Output:

`Sum of left leaves is 78`

Time complexity of the above solution is O(n) where n is number of nodes in Binary Tree.

Following is Another Method to solve the above problem. This solution passes in a sum variable as an accumulator. When a left leaf is encountered, the leaf’s data is added to sum. Time complexity of this method is also O(n). Thanks to Xin Tong (geeksforgeeks userid trent.tong) for suggesting this method.

## C++

 `// A C++ program to find sum of all left leaves ` `#include ` `using` `namespace` `std; ` ` `  `/* A binary tree Node has key, pointer to left and right ` `   ``children */` `struct` `Node ` `{ ` `    ``int` `key; ` `    ``struct` `Node* left, *right; ` `}; ` ` `  `/* Helper function that allocates a new node with the ` `   ``given data and NULL left and right pointer. */` `Node *newNode(``char` `k) ` `{ ` `    ``Node *node = ``new` `Node; ` `    ``node->key = k; ` `    ``node->right = node->left = NULL; ` `    ``return` `node; ` `} ` ` `  `/* Pass in a sum variable as an accumulator */` `void` `leftLeavesSumRec(Node *root, ``bool` `isleft, ``int` `*sum) ` `{ ` `    ``if` `(!root) ``return``; ` ` `  `    ``// Check whether this node is a leaf node and is left. ` `    ``if` `(!root->left && !root->right && isleft) ` `        ``*sum += root->key; ` ` `  `    ``// Pass 1 for left and 0 for right ` `    ``leftLeavesSumRec(root->left,  1, sum); ` `    ``leftLeavesSumRec(root->right, 0, sum); ` `} ` ` `  `// A wrapper over above recursive function ` `int` `leftLeavesSum(Node *root) ` `{ ` `    ``int` `sum = 0; ``//Initialize result ` ` `  `    ``// use the above recursive function to evaluate sum ` `    ``leftLeavesSumRec(root, 0, &sum); ` ` `  `    ``return` `sum; ` `} ` ` `  `/* Driver program to test above functions*/` `int` `main() ` `{ ` `    ``// Let us construct the Binary Tree shown in the ` `    ``// above figure ` `    ``int` `sum = 0; ` `    ``struct` `Node *root         = newNode(20); ` `    ``root->left                = newNode(9); ` `    ``root->right               = newNode(49); ` `    ``root->right->left         = newNode(23); ` `    ``root->right->right        = newNode(52); ` `    ``root->right->right->left  = newNode(50); ` `    ``root->left->left          = newNode(5); ` `    ``root->left->right         = newNode(12); ` `    ``root->left->right->right  = newNode(12); ` ` `  `    ``cout << ``"Sum of left leaves is "` `<< leftLeavesSum(root) << endl; ` `    ``return` `0; ` `} `

## Java

 `// Java program to find sum of all left leaves ` `class` `Node  ` `{ ` `    ``int` `data; ` `    ``Node left, right; ` `  `  `    ``Node(``int` `item) { ` `        ``data = item; ` `        ``left = right = ``null``; ` `    ``} ` `} ` `  `  `// Passing sum as accumulator and implementing pass by reference  ` `// of sum variable  ` `class` `Sum  ` `{ ` `    ``int` `sum = ``0``; ` `} ` `  `  `class` `BinaryTree  ` `{ ` `    ``Node root; ` `  `  `    ``/* Pass in a sum variable as an accumulator */` `    ``void` `leftLeavesSumRec(Node node, ``boolean` `isleft, Sum summ)  ` `    ``{ ` `        ``if` `(node == ``null``) ` `            ``return``; ` `  `  `        ``// Check whether this node is a leaf node and is left. ` `        ``if` `(node.left == ``null` `&& node.right == ``null` `&& isleft) ` `            ``summ.sum = summ.sum + node.data; ` `  `  `        ``// Pass true for left and false for right ` `        ``leftLeavesSumRec(node.left, ``true``, summ); ` `        ``leftLeavesSumRec(node.right, ``false``, summ); ` `    ``} ` `  `  `    ``// A wrapper over above recursive function ` `    ``int` `leftLeavesSum(Node node)  ` `    ``{ ` `        ``Sum suum = ``new` `Sum(); ` `         `  `        ``// use the above recursive function to evaluate sum ` `        ``leftLeavesSumRec(node, ``false``, suum); ` `  `  `        ``return` `suum.sum; ` `    ``} ` `  `  `    ``// Driver program ` `    ``public` `static` `void` `main(String args[])  ` `    ``{ ` `        ``BinaryTree tree = ``new` `BinaryTree(); ` `        ``tree.root = ``new` `Node(``20``); ` `        ``tree.root.left = ``new` `Node(``9``); ` `        ``tree.root.right = ``new` `Node(``49``); ` `        ``tree.root.left.right = ``new` `Node(``12``); ` `        ``tree.root.left.left = ``new` `Node(``5``); ` `        ``tree.root.right.left = ``new` `Node(``23``); ` `        ``tree.root.right.right = ``new` `Node(``52``); ` `        ``tree.root.left.right.right = ``new` `Node(``12``); ` `        ``tree.root.right.right.left = ``new` `Node(``50``); ` `  `  `        ``System.out.println(``"The sum of leaves is "` `+  ` `                                    ``tree.leftLeavesSum(tree.root)); ` `    ``} ` `} ` `  `  `// This code is contributed by Mayank Jaiswal  `

## Python

 `# Python program to find sum of all left leaves ` ` `  `# A binary tree node ` `class` `Node: ` ` `  `    ``# A constructor to create a new Node ` `    ``def` `__init__(``self``, key): ` `        ``self``.key ``=` `key ` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None` ` `  `def` `leftLeavesSumRec(root, isLeft, summ): ` `    ``if` `root ``is` `None``: ` `        ``return` `     `  `    ``# Check whether this node is a leaf node and is left ` `    ``if` `root.left ``is` `None` `and` `root.right ``is` `None` `and` `isLeft ``=``=` `True``: ` `        ``summ[``0``] ``+``=` `root.key ` ` `  `    ``# Pass 1 for left and 0 for right ` `    ``leftLeavesSumRec(root.left, ``1``, summ) ` `    ``leftLeavesSumRec(root.right, ``0``, summ) ` `     `  ` `  `# A wrapper over above recursive function ` `def` `leftLeavesSum(root): ` `    ``summ ``=` `[``0``] ``# initialize result ` `     `  `    ``# Use the above recursive fucntion to evaluate sum ` `    ``leftLeavesSumRec(root, ``0``, summ) ` `     `  `    ``return` `summ[``0``] ` ` `  `# Driver program to test above function ` ` `  `# Let us construct the Binary Tree shown in the ` `# above figure ` `root ``=` `Node(``20``); ` `root.left``=` `Node(``9``); ` `root.right   ``=` `Node(``49``); ` `root.right.left ``=` `Node(``23``); ` `root.right.right``=` `Node(``52``); ` `root.right.right.left  ``=` `Node(``50``); ` `root.left.left  ``=` `Node(``5``); ` `root.left.right ``=` `Node(``12``); ` `root.left.right.right  ``=` `Node(``12``); ` ` `  `print` `"Sum of left leaves is"``, leftLeavesSum(root)   ` ` `  `# This code is contributed by Nikhil Kumar Singh(nickzuck_007) `

## C#

 `using` `System; ` ` `  `// C# program to find sum of all left leaves  ` `public` `class` `Node ` `{ ` `    ``public` `int` `data; ` `    ``public` `Node left, right; ` ` `  `    ``public` `Node(``int` `item) ` `    ``{ ` `        ``data = item; ` `        ``left = right = ``null``; ` `    ``} ` `} ` ` `  `// Passing sum as accumulator and implementing pass by reference   ` `// of sum variable   ` `public` `class` `Sum ` `{ ` `    ``public` `int` `sum = 0; ` `} ` ` `  `public` `class` `BinaryTree ` `{ ` `    ``public` `Node root; ` ` `  `    ``/* Pass in a sum variable as an accumulator */` `    ``public` `virtual` `void` `leftLeavesSumRec(Node node, ``bool` `isleft, Sum summ) ` `    ``{ ` `        ``if` `(node == ``null``) ` `        ``{ ` `            ``return``; ` `        ``} ` ` `  `        ``// Check whether this node is a leaf node and is left.  ` `        ``if` `(node.left == ``null` `&& node.right == ``null` `&& isleft) ` `        ``{ ` `            ``summ.sum = summ.sum + node.data; ` `        ``} ` ` `  `        ``// Pass true for left and false for right  ` `        ``leftLeavesSumRec(node.left, ``true``, summ); ` `        ``leftLeavesSumRec(node.right, ``false``, summ); ` `    ``} ` ` `  `    ``// A wrapper over above recursive function  ` `    ``public` `virtual` `int` `leftLeavesSum(Node node) ` `    ``{ ` `        ``Sum suum = ``new` `Sum(); ` ` `  `        ``// use the above recursive function to evaluate sum  ` `        ``leftLeavesSumRec(node, ``false``, suum); ` ` `  `        ``return` `suum.sum; ` `    ``} ` ` `  `    ``// Driver program  ` `    ``public` `static` `void` `Main(``string``[] args) ` `    ``{ ` `        ``BinaryTree tree = ``new` `BinaryTree(); ` `        ``tree.root = ``new` `Node(20); ` `        ``tree.root.left = ``new` `Node(9); ` `        ``tree.root.right = ``new` `Node(49); ` `        ``tree.root.left.right = ``new` `Node(12); ` `        ``tree.root.left.left = ``new` `Node(5); ` `        ``tree.root.right.left = ``new` `Node(23); ` `        ``tree.root.right.right = ``new` `Node(52); ` `        ``tree.root.left.right.right = ``new` `Node(12); ` `        ``tree.root.right.right.left = ``new` `Node(50); ` ` `  `        ``Console.WriteLine(``"The sum of leaves is "` `+ tree.leftLeavesSum(tree.root)); ` `    ``} ` `} ` ` `  `  ``// This code is contributed by Shrikant13 `

Output:

`Sum of left leaves is 78`

Iterative Approach :
This is the Iterative Way to find the sum of the left leaves.
Idea is to perform Depth-First Traversal on the tree (either Inorder, Preorder or Postorder) using a stack and checking if the Left Child is a Leaf node. If it is, then add the nodes value to the sum variable

## C++

 `// C++ program to find sum of all left leaves ` `#include ` `using` `namespace` `std; ` ` `  `// A binary tree node ` `class` `Node ` `{ ` `    ``public``: ` `    ``int` `key; ` `    ``Node* left, *right; ` `     `  `    ``// A constructor to create a new Node ` `    ``Node(``int` `key_) ` `    ``{ ` `        ``key = key_; ` `        ``left = NULL; ` `        ``right = NULL; ` `    ``} ` `}; ` ` `  `// Return the sum of left leaf nodes ` `int` `sumOfLeftLeaves(Node* root) ` `{ ` `    ``if``(root == NULL) ` `        ``return` `0; ` `     `  `    ``// Using a stack_ for Depth-First Traversal of the tree ` `    ``stack stack_;  ` `    ``stack_.push(root); ` `     `  `    ``// sum holds the sum of all the left leaves ` `    ``int` `sum = 0; ` ` `  `    ``while``(stack_.size() > 0) ` `    ``{ ` `        ``Node* currentNode = stack_.top(); ` `        ``stack_.pop(); ` ` `  `        ``if` `(currentNode->left != NULL) ` `        ``{ ` `            ``stack_.push(currentNode->left); ` `             `  `            ``// Check if currentNode's left child is a leaf node ` `            ``if``(currentNode->left->left == NULL && currentNode->left->right == NULL) ` `            ``{ ` `                ``// if currentNode is a leaf, add its data to the sum  ` `                ``sum = sum + currentNode->left->key ; ` `            ``} ` `        ``} ` `        ``if` `(currentNode->right != NULL) ` `            ``stack_.push(currentNode->right); ` `    ``} ` `         `  `    ``return` `sum; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``Node *root = ``new` `Node(20); ` `    ``root->left= ``new` `Node(9); ` `    ``root->right = ``new` `Node(49); ` `    ``root->right->left = ``new` `Node(23); ` `    ``root->right->right= ``new` `Node(52); ` `    ``root->right->right->left = ``new` `Node(50); ` `    ``root->left->left = ``new` `Node(5); ` `    ``root->left->right = ``new` `Node(12); ` `    ``root->left->right->right = ``new` `Node(12); ` `     `  `    ``cout << ``"Sum of left leaves is "` `<< sumOfLeftLeaves(root) << endl; ` `    ``return` `0; ` `} ` ` `  `// This code is contributed by Arnab Kundu `

## Python3

 `# Python3 program to find sum of all left leaves ` ` `  `# A binary tree node ` `class` `Node: ` ` `  `    ``# A constructor to create a new Node ` `    ``def` `__init__(``self``, key): ` `        ``self``.key ``=` `key ` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None` ` `  ` `  `# Return the sum of left leaf nodes ` `def` `sumOfLeftLeaves(root): ` `    ``if``(root ``is` `None``): ` `        ``return` `     `  `    ``# Using a stack for Depth-First Traversal of the tree ` `    ``stack ``=` `[]    ` `    ``stack.append(root) ` `     `  `    ``# sum holds the sum of all the left leaves ` `    ``sum` `=` `0` ` `  `    ``while` `len``(stack) > ``0``: ` `        ``currentNode ``=` `stack.pop() ` ` `  `        ``if` `currentNode.left ``is` `not` `None``: ` `            ``stack.append(currentNode.left) ` `             `  `            ``# Check if currentNode's left child is a leaf node ` `            ``if` `currentNode.left.left ``is` `None` `and` `currentNode.left.right ``is` `None``: ` ` `  `                ``# if currentNode is a leaf, add its data to the sum   ` `                ``sum` `=` `sum` `+` `currentNode.left.data  ` ` `  `        ``if` `currentNode.right ``is` `not` `None``: ` `            ``stack.append(currentNode.right) ` `    ``return` `sum` ` `  `# Driver Code ` `root ``=` `Tree(``20``); ` `root.left``=` `Tree(``9``); ` `root.right   ``=` `Tree(``49``); ` `root.right.left ``=` `Tree(``23``); ` `root.right.right``=` `Tree(``52``); ` `root.right.right.left  ``=` `Tree(``50``); ` `root.left.left  ``=` `Tree(``5``); ` `root.left.right ``=` `Tree(``12``); ` `root.left.right.right  ``=` `Tree(``12``); ` ` `  `print``(``'Sum of left leaves is {}'``.``format``(sumOfLeftLeaves(root))) `

Output:

`Sum of left leaves is 78`

Thanks to Shubham Tambere for suggesting this approach.

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