Root to leaf path sum equal to a given number

• Difficulty Level : Medium
• Last Updated : 10 Jan, 2022

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 childand 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 rootdown to a leaf, such that adding up all the values along the pathequals 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 when you reach the leaf node.*/bool hasPathSum(node* Node, int sum){          bool ans = 0;   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;   /* otherwise check both subtrees */  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 thegiven 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 Codeint 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 reach the leaf node.*/bool hasPathSum(struct node* node, int sum){     bool ans = 0;   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;   /* otherwise check both subtrees */  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 Codeint 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// 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 you reach the leaf node.     */     boolean hasPathSum(Node node, int sum)    {      boolean ans = false;      int subSum = sum - node.data;      if(subSum == 0 && node.left == null && node.right == null)        return(ans = true);      if(node.left != null)                  // ans || hasPathSum... has no utility if the ans is false        ans = ans || hasPathSum(node.left, subSum);                   if(node.right != null)                 // But if it is true then we can avoid calling hasPathSum        // here as answer has already been found        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        */        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)

Python3

 # 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):        ans = 0        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         # Otherwise check both subtrees        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 Code  s = 21root = 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 root to// leaf path sum equal to a given numberusing System; /* A binary tree node has data, pointerto 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 you reach the leaf node..    */     public virtual bool haspathSum(Node node, int sum)    {      ans = false;       int subsum = sum - node.data;      if (subsum == 0 && node.left == null          && node.right == null) {        return true;      }       /* otherwise check both subtrees */      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

Javascript


Output
There is a root-to-leaf path with sum 21

Time Complexity: O(n)
References: http://cslibrary.stanford.edu/110/BinaryTrees.html
Author: Tushar Roy

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

My Personal Notes arrow_drop_up