# Shortest root to leaf path sum equal to a given number

Given a binary tree and a number, the task is to return the length of the shortest path beginning at root and ending at a leaf node such that the sum of numbers along that path is equal to ‘sum’. Print “-1” if no such path exists.

**Examples:**

Input:1 / \ 10 15 / \ 5 2 Number = 16Output:2 There are two paths: 1->15, length of path is 3 1->10->5, length of path is 2Input:1 / \ 10 15 / \ 5 2 Number = 20Output:-1 There exists no such path with 20 as sum from root to any node

**Source:** Microsoft Interview

**Approach**: An approach to check whether such a path exists or not has been discussed in this post. The below steps can be followed to find the shortest path.

- Recursively call the left and right child with path sum and level to ( number – value at the current node, level+1).
- If the leaf node is reached, then store the minimum of all the levels of the leaf node.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the above approach ` `#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; ` ` ` `struct` `Node* right; ` `}; ` ` ` `// Function to find the shortes path from root ` `// to leaf with given sum ` `void` `hasPathSum(` `struct` `Node* node, ` `int` `sum, ` ` ` `int` `& mini, ` `int` `level) ` `{ ` ` ` `// check if leaf node and check sum ` ` ` `if` `(node == NULL) { ` ` ` ` ` `// Store the minimum path ` ` ` `if` `(sum == 0) ` ` ` `mini = min(level - 1, mini); ` ` ` ` ` `return` `; ` ` ` `} ` ` ` `else` `{ ` ` ` `int` `subSum = sum - (node->data); ` ` ` ` ` `// Recur in the left tree ` ` ` `hasPathSum(node->left, subSum, mini, level + 1); ` ` ` ` ` `// Recur in the right tree ` ` ` `hasPathSum(node->right, subSum, mini, level + 1); ` ` ` `} ` `} ` ` ` `/* 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 = ` `new` `Node; ` ` ` `node->data = data; ` ` ` `node->left = NULL; ` ` ` `node->right = NULL; ` ` ` ` ` `return` `(node); ` `} ` ` ` `/* Driver program to test above functions*/` `int` `main() ` `{ ` ` ` ` ` `int` `sum = 16; ` ` ` ` ` `/* Constructed binary tree is ` ` ` ` ` `1 ` ` ` `/ \ ` ` ` `10 15 ` ` ` `/ \ ` ` ` `5 2 ` `*/` ` ` `struct` `Node* root = newnode(1); ` ` ` `root->left = newnode(10); ` ` ` `root->right = newnode(15); ` ` ` `root->left->left = newnode(5); ` ` ` `root->left->right = newnode(2); ` ` ` ` ` `int` `mini = INT_MAX; ` ` ` ` ` `// Function call ` ` ` `hasPathSum(root, sum, mini, 0); ` ` ` ` ` `if` `(mini != INT_MAX) ` ` ` `printf` `(` `"There is a root-to-leaf path with sum %d\n"` ` ` `"and the shortest path length is: %d"` `, ` ` ` `sum, mini); ` ` ` `else` ` ` `printf` `(` `"There is no root-to-leaf path with sum %d"` `, sum); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## Python3

# Python3 implementation of the above approach

INT_MAX = 2147483647

“”” A binary tree node has data, pointer

to left child and a pointer to right child “””

“””UTILITY FUNCTIONS

Helper function that allocates a new node

with the given data and NULL left and right

pointers.”””

class newnode:

# Construct to create a newNode

def __init__(self, key):

self.data = key

self.left = None

self.right = None

# Function to find the shortes path

# from root to leaf with given summ

def hasPathSum(node, summ, mini, level) :

# check if leaf node and check summ

if (node == None) :

# Store the minimum path

if (summ[0] == 0) :

mini[0] = min(level – 1, mini[0])

return

else:

subsumm = [summ[0] – (node.data)]

# Recur in the left tree

hasPathSum(node.left, subsumm,

mini, level + 1)

# Recur in the right tree

hasPathSum(node.right, subsumm,

mini, level + 1)

# Driver Code

if __name__ == ‘__main__’:

summ = [16]

“”” Constructed binary tree is

1

/ \

10 15

/ \

5 2

“””

root = newnode(1)

root.left = newnode(10)

root.right = newnode(15)

root.left.left = newnode(5)

root.left.right = newnode(2)

mini = [INT_MAX]

# Function call

hasPathSum(root, summ, mini, 0)

if (mini != INT_MAX) :

print(“There is a root-to-leaf path”,

“with sum”, summ[0])

print(“and the shortest path length is:”,

mini[0])

else:

print(“There is no root-to-leaf path”,

“with sum “, summ)

# This code is contributed by

# Shubham Singh(SHUBHAMSINGH10)

**Output:**

There is a root-to-leaf path with sum 16 and the shortest path length is: 1

## Recommended Posts:

- Root to leaf path sum equal to a given number
- Root to leaf path sum equal to a given number in BST
- Root to leaf path product equal to a given number
- Print the first shortest root to leaf path in a Binary Tree
- Find if there is a pair in root to a leaf path with sum equals to root's data
- Check if there is a root to leaf path with given sequence
- Sum of nodes on the longest path from root to leaf node
- Root to leaf path with maximum distinct nodes
- Find the maximum sum leaf to root path in a Binary Tree
- Root to leaf paths having equal lengths in a Binary Tree
- Print the longest leaf to leaf path in a Binary tree
- Sum of all the numbers that are formed from root to leaf paths
- Print root to leaf paths without using recursion
- Given a binary tree, print all root-to-leaf paths
- Print all root to leaf paths with there relative positions

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