# Root to leaf path sum equal to a given number

Given a binary tree and a number, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals the given number. Return false if no such path can be found. For example, in the above tree root to leaf paths exist with following sums.

21 –> 10 – 8 – 3
23 –> 10 – 8 – 5
14 –> 10 – 2 – 2

So the returned value should be true only for numbers 21, 23 and 14. For any other number, returned value should be false.

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Algorithm:
Recursively check if left or right child has path sum equal to ( number – value at current node)

Implementation:

## C++

 `#include ` `using` `namespace` `std; ` `#define bool int  ` ` `  `/* A binary tree node has data, pointer to left child  ` `and a pointer to right child */` `class` `node  ` `{  ` `    ``public``: ` `    ``int` `data;  ` `    ``node* left;  ` `    ``node* right;  ` `};  ` ` `  `/*  ` `Given a tree and a sum, return true if there is a path from the root  ` `down to a leaf, such that adding up all the values along the path  ` `equals the given sum.  ` ` `  `Strategy: subtract the node value from the sum when recurring down,  ` `and check to see if the sum is 0 when you run out of tree.  ` `*/` `bool` `hasPathSum(node* Node, ``int` `sum)  ` `{  ` `    ``/* return true if we run out of tree and sum==0 */` `    ``if` `(Node == NULL)  ` `    ``{  ` `        ``return` `(sum == 0);  ` `    ``}  ` `     `  `    ``else` `    ``{  ` `        ``bool` `ans = 0;  ` `     `  `        ``/* otherwise check both subtrees */` `        ``int` `subSum = sum - Node->data;  ` `     `  `        ``/* If we reach a leaf node and sum becomes 0 then return true*/` `        ``if` `( subSum == 0 && Node->left == NULL && Node->right == NULL )  ` `        ``return` `1;  ` `     `  `        ``if``(Node->left)  ` `            ``ans = ans || hasPathSum(Node->left, subSum);  ` `        ``if``(Node->right)  ` `            ``ans = ans || hasPathSum(Node->right, subSum);  ` `     `  `        ``return` `ans;  ` `    ``}  ` `}  ` ` `  `/* UTILITY FUNCTIONS */` `/* Helper function that allocates a new node with the  ` `given data and NULL left and right pointers. */` `node* newnode(``int` `data)  ` `{  ` `    ``node* Node = ``new` `node(); ` `    ``Node->data = data;  ` `    ``Node->left = NULL;  ` `    ``Node->right = NULL;  ` `     `  `    ``return``(Node);  ` `}  ` ` `  `/* Driver program to test above functions*/` `int` `main()  ` `{  ` ` `  `    ``int` `sum = 21;  ` `     `  `    ``/* Constructed binary tree is  ` `                ``10  ` `            ``/ \  ` `            ``8 2  ` `        ``/ \ /  ` `        ``3 5 2  ` `    ``*/` `    ``node *root = newnode(10);  ` `    ``root->left = newnode(8);  ` `    ``root->right = newnode(2);  ` `    ``root->left->left = newnode(3);  ` `    ``root->left->right = newnode(5);  ` `    ``root->right->left = newnode(2);  ` `     `  `    ``if``(hasPathSum(root, sum))  ` `        ``cout << ``"There is a root-to-leaf path with sum "` `<< sum;  ` `    ``else` `        ``cout << ``"There is no root-to-leaf path with sum "` `<< sum;  ` `     `  `    ``return` `0;  ` `}  ` ` `  `// This code is contributed by rathbhupendra `

## C

 `#include ` `#include ` `#define bool int ` ` `  `/* A binary tree node has data, pointer to left child ` `   ``and a pointer to right child */` `struct` `node ` `{ ` `   ``int` `data; ` `   ``struct` `node* left; ` `   ``struct` `node* right; ` `}; ` ` `  `/* ` ` ``Given a tree and a sum, return true if there is a path from the root ` ` ``down to a leaf, such that adding up all the values along the path ` ` ``equals the given sum. ` ` `  ` ``Strategy: subtract the node value from the sum when recurring down, ` ` ``and check to see if the sum is 0 when you run out of tree. ` `*/` `bool` `hasPathSum(``struct` `node* node, ``int` `sum) ` `{ ` `  ``/* return true if we run out of tree and sum==0 */` `  ``if` `(node == NULL) ` `  ``{ ` `     ``return` `(sum == 0); ` `  ``} ` `  `  `  ``else` `  ``{ ` `    ``bool` `ans = 0;   ` `  `  `    ``/* otherwise check both subtrees */` `    ``int` `subSum = sum - node->data; ` `  `  `    ``/* If we reach a leaf node and sum becomes 0 then return true*/` `    ``if` `( subSum == 0 && node->left == NULL && node->right == NULL ) ` `      ``return` `1; ` `  `  `    ``if` `(node->left) ` `      ``ans = ans || hasPathSum(node->left, subSum); ` `    ``if` `(node->right) ` `      ``ans = ans || hasPathSum(node->right, subSum); ` `  `  `    ``return` `ans; ` `  ``} ` `} ` ` `  `/* UTILITY FUNCTIONS */` `/* 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 functions*/` `int` `main() ` `{ ` ` `  `  ``int` `sum = 21; ` ` `  `  ``/* Constructed binary tree is ` `            ``10 ` `          ``/   \ ` `        ``8      2 ` `      ``/  \    / ` `    ``3     5  2 ` `  ``*/` `  ``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->left = newnode(2); ` ` `  `  ``if``(hasPathSum(root, sum)) ` `    ``printf``(``"There is a root-to-leaf path with sum %d"``, sum); ` `  ``else` `    ``printf``(``"There is no root-to-leaf path with sum %d"``, sum); ` ` `  `  ``getchar``(); ` `  ``return` `0; ` `} `

## Java

 `// Java program to print to print root to leaf path sum equal to ` `// a given number ` `  `  `/* A binary tree node has data, pointer to left child ` `   ``and a pointer to right child */` `class` `Node  ` `{ ` `    ``int` `data; ` `    ``Node left, right; ` `  `  `    ``Node(``int` `item)  ` `    ``{ ` `        ``data = item; ` `        ``left = right = ``null``; ` `    ``} ` `} ` `  `  `class` `BinaryTree { ` `  `  `    ``Node root; ` `      `  `    ``/* ` `     ``Given a tree and a sum, return true if there is a path from the root ` `     ``down to a leaf, such that adding up all the values along the path ` `     ``equals the given sum. ` `   `  `     ``Strategy: subtract the node value from the sum when recurring down, ` `     ``and check to see if the sum is 0 when you run out of tree. ` `     ``*/` `  `  `    ``boolean` `hasPathSum(Node node, ``int` `sum)  ` `    ``{ ` `        ``if` `(node == ``null``) ` `            ``return` `sum == ``0``; ` `        ``return` `hasPathSum(node.left, sum - node.data) ||  ` `               ``hasPathSum(node.right, sum - node.data); ` `    ``} ` `     `  `    ``/* Driver program to test the above functions */` `    ``public` `static` `void` `main(String args[])  ` `    ``{ ` `        ``int` `sum = ``21``; ` `         `  `        ``/* Constructed binary tree is ` `              ``10 ` `             ``/  \ ` `           ``8     2 ` `          ``/ \   / ` `         ``3   5 2 ` `        ``*/` `        ``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.left = ``new` `Node(``2``); ` `  `  `        ``if` `(tree.haspathSum(tree.root, sum)) ` `            ``System.out.println(``"There is a root to leaf path with sum "` `+ sum); ` `        ``else` `            ``System.out.println(``"There is no root to leaf path with sum "` `+ sum); ` `    ``} ` `} ` `  `  `// This code has been contributed by Mayank Jaiswal(mayank_24) `

## Python

 `# Python program to find if there is a root to sum path ` ` `  `# A binary tree node  ` `class` `Node: ` ` `  `    ``# Constructor to create a new node ` `    ``def` `__init__(``self``, data): ` `        ``self``.data ``=` `data  ` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None` ` `  `"""  ` ` ``Given a tree and a sum, return true if there is a path from the root ` ` ``down to a leaf, such that adding up all the values along the path ` ` ``equals the given sum. ` `  `  ` ``Strategy: subtract the node value from the sum when recurring down, ` ` ``and check to see if the sum is 0 when you run out of tree. ` `"""` `# s is the sum ` `def` `hasPathSum(node, s): ` `     `  `    ``# Return true if we run out of tree and s = 0  ` `    ``if` `node ``is` `None``: ` `        ``return` `(s ``=``=` `0``) ` ` `  `    ``else``: ` `        ``ans ``=` `0`  `         `  `        ``# Otherwise check both subtrees ` `        ``subSum ``=` `s ``-` `node.data  ` `         `  `        ``# If we reach a leaf node and sum becomes 0, then  ` `        ``# return True  ` `        ``if``(subSum ``=``=` `0` `and` `node.left ``=``=` `None` `and` `node.right ``=``=` `None``): ` `            ``return` `True`  ` `  `        ``if` `node.left ``is` `not` `None``: ` `            ``ans ``=` `ans ``or` `hasPathSum(node.left, subSum) ` `        ``if` `node.right ``is` `not` `None``: ` `            ``ans ``=` `ans ``or` `hasPathSum(node.right, subSum) ` ` `  `        ``return` `ans  ` ` `  `# Driver program to test above functions ` ` `  `s ``=` `21` `root ``=` `Node(``10``) ` `root.left ``=` `Node(``8``) ` `root.right ``=` `Node(``2``) ` `root.left.right ``=` `Node(``5``) ` `root.left.left ``=` `Node(``3``) ` `root.right.left ``=` `Node(``2``) ` ` `  `if` `hasPathSum(root, s): ` `    ``print` `"There is a root-to-leaf path with sum %d"` `%``(s) ` `else``: ` `    ``print` `"There is no root-to-leaf path with sum %d"` `%``(s) ` ` `  `# This code is contributed by Nikhil Kumar Singh(nickzuck_007) `

## C#

 `// C# program to print to print root to  ` `// leaf path sum equal to a given number  ` `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` `item) ` `    ``{ ` `        ``data = item; ` `        ``left = right = ``null``; ` `    ``} ` `} ` ` `  `class` `GFG ` `{ ` `public` `Node root; ` ` `  `/*  ` `Given a tree and a sum, return true if  ` `there is a path from the root down to a  ` `leaf, such that adding up all the values   ` `along the path equals the given sum.  ` ` `  `Strategy: subtract the node value from the  ` `sum when recurring down, and check to see ` `if the sum is 0 when you run out of tree.  ` `*/` ` `  `public` `virtual` `bool` `haspathSum(Node node,  ` `                               ``int` `sum) ` `{ ` `    ``if` `(node == ``null``) ` `    ``{ ` `        ``return` `(sum == 0); ` `    ``} ` `    ``else` `    ``{ ` `        ``bool` `ans = ``false``; ` ` `  `        ``/* otherwise check both subtrees */` `        ``int` `subsum = sum - node.data; ` `        ``if` `(subsum == 0 && node.left == ``null` `&&  ` `                           ``node.right == ``null``) ` `        ``{ ` `            ``return` `true``; ` `        ``} ` `        ``if` `(node.left != ``null``) ` `        ``{ ` `            ``ans = ans || haspathSum(node.left, subsum); ` `        ``} ` `        ``if` `(node.right != ``null``) ` `        ``{ ` `            ``ans = ans || haspathSum(node.right, subsum); ` `        ``} ` `        ``return` `ans; ` `    ``} ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(``string``[] args) ` `{ ` `    ``int` `sum = 21; ` ` `  `    ``/* Constructed binary tree is  ` `        ``10  ` `        ``/ \  ` `    ``8     2  ` `    ``/ \ /  ` `    ``3 5 2  ` `    ``*/` `    ``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.left = ``new` `Node(2); ` ` `  `    ``if` `(tree.haspathSum(tree.root, sum)) ` `    ``{ ` `        ``Console.WriteLine(``"There is a root to leaf "` `+ ` `                              ``"path with sum "` `+ sum); ` `    ``} ` `    ``else` `    ``{ ` `        ``Console.WriteLine(``"There is no root to leaf "` `+  ` `                               ``"path with sum "` `+ sum); ` `    ``} ` `} ` `} ` ` `  `// This code is contributed by Shrikant13 `

Output :

`There is a root to leaf path with sum 21`

Time Complexity: O(n)

Author: Tushar Roy

Please write comments if you find any bug in above code/algorithm, or find other ways to solve the same problem

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