Given a binary tree and a number, the task is to return the length of the shortest path beginning at the 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 = 16
Output: 2
There are two paths:
1->15, length of path is 3
1->10->5, length of path is 2
Input:
1
/ \
10 15
/ \
5 2
Number = 20
Output: -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 sumvoid hasPathSum(struct Node* node, int sum, int& mini, int level){ // went beyond tree if (node == NULL) return; int subSum = sum - (node->data); if (node->left == NULL && node->right == NULL) { // Store the minimum path if (subSum == 0) mini = min(level - 1, mini); return; } else { // 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 + 1); else printf("There is no root-to-leaf path with sum %d", sum); return 0;} |
Java
// Java implementation of the above approach class GFG {static int mini;/* A binary tree node has data, pointer to left child and a pointer to right child */static class Node{ int data; Node left; Node right; }; // Function to find the shortes path from root // to leaf with given sum static void hasPathSum(Node node, int sum, int level) { // went beyond tree if (node == null) return; int subSum = sum - (node.data); if (node.left == null && node.right == null) { // Store the minimum path if (subSum == 0) mini = Math.min(level - 1, mini); return; } else { // Recur in the left tree hasPathSum(node.left, subSum, level + 1); // Recur in the right tree hasPathSum(node.right, subSum, level + 1); } } /* UTILITY FUNCTIONS *//* Helper function that allocates a new node with the given data and null left and right pointers. */static Node newnode(int data) { Node node = new Node(); node.data = data; node.left = null; node.right = null; return (node); } /* Driver code*/public static void main(String[] args){ int sum = 16; /* Constructed binary tree is 1 / \ 10 15 / \ 5 2 */ Node root = newnode(1); root.left = newnode(10); root.right = newnode(15); root.left.left = newnode(5); root.left.right = newnode(2); mini = Integer.MAX_VALUE; // Function call hasPathSum(root, sum, 0); if (mini != Integer.MAX_VALUE) System.out.printf("There is a root-to-leaf path with sum %d\n" +"and the shortest path length is: %d" ,sum, mini + 1); else System.out.printf("There is no root-to-leaf path with sum %d", sum); }}// This code is contributed by Princi Singh |
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) : return subsumm = summ - node.data if(node.left == None and node.right == None): # Store the minimum path if (subsumm == 0) : mini[0] = min(level - 1, mini[0]) return else: # 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 15 """ 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[0] != INT_MAX) : print("There is a root-to-leaf path", "with sum", summ) print("and the shortest path length is:", mini[0] + 1) else: print("There is no root-to-leaf path", "with sum ", summ) # This code is contributed by # Shubham Singh(SHUBHAMSINGH10) |
C#
// C# implementation of the above approachusing System; class GFG {static int mini;/* 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; public Node right; }; // Function to find the shortes path from root // to leaf with given sum static void hasPathSum(Node node, int sum, int level) { // went beyond tree if (node == null) return; int subSum = sum - (node.data); if (node.left == null && node.right == null) { // Store the minimum path if (subSum == 0) mini = Math.Min(level - 1, mini); return; } else { // Recur in the left tree hasPathSum(node.left, subSum, level + 1); // Recur in the right tree hasPathSum(node.right, subSum, level + 1); } } /* UTILITY FUNCTIONS *//* Helper function that allocates a new node with the given data and null left and right pointers. */static Node newnode(int data) { Node node = new Node(); node.data = data; node.left = null; node.right = null; return (node); } /* Driver code*/public static void Main(String[] args){ int sum = 16; /* Constructed binary tree is 1 / \ 10 15 / \ 5 2 */ Node root = newnode(1); root.left = newnode(10); root.right = newnode(15); root.left.left = newnode(5); root.left.right = newnode(2); mini = int.MaxValue; // Function call hasPathSum(root, sum, 0); if (mini != int.MaxValue) Console.Write("There is a root-to-leaf path with sum {0}\n" +"and the shortest path length is: {1}" ,sum, mini + 1); else Console.Write("There is no root-to-leaf path with sum {0}", sum); }}// This code is contributed by Rajput-Ji |
Javascript
<script>// JavaScript implementation of the // above approach /* * A binary tree node has data, pointer to left child and a pointer to right * child */class Node { constructor() { this.data = 0; this.left = null; this.right = null; } } // Function to find the shortes path from root // to leaf with given sum function hasPathSum(node , sum , level) { // went beyond tree if (node == null) return; var subSum = sum - (node.data); if (node.left == null && node.right == null) { // Store the minimum path if (subSum == 0) mini = Math.min(level - 1, mini); return; } else { // Recur in the left tree hasPathSum(node.left, subSum, level + 1); // Recur in the right tree hasPathSum(node.right, subSum, level + 1); } } /* UTILITY FUNCTIONS */ /* * Helper function that allocates a new node with the given data and null left * and right pointers. */ function newnode(data) { var node = new Node(); node.data = data; node.left = null; node.right = null; return (node); } /* Driver code */ var sum = 16; /* * Constructed binary tree is * * 1 / \ 10 15 / \ 5 2 */ var root = newnode(1); root.left = newnode(10); root.right = newnode(15); root.left.left = newnode(5); root.left.right = newnode(2); mini = Number.MAX_VALUE; // Function call hasPathSum(root, sum, 0); if (mini != Number.MAX_VALUE) document.write( "There is a root-to-leaf path with sum "+sum+"<br/>" + "and the shortest path length is: "+(mini + 1)); else document.write( "There is no root-to-leaf path with sum "+ sum );// This code is contributed by todaysgaurav </script> |
Output
There is a root-to-leaf path with sum 16 and the shortest path length is: 1
Complexity Analysis:
- Time Complexity: O(N), as we are using recursion to traverse all the nodes of the tree. Where N is the number of nodes in the tree.
- Auxiliary Space: O(1), as we are not using any extra space.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
… [Trackback]
[…] Info on that Topic: geeksforgeeks.org/shortest-root-to-leaf-path-sum-equal-to-a-given-number/ […]
… [Trackback]
[…] Read More Info here on that Topic: geeksforgeeks.org/shortest-root-to-leaf-path-sum-equal-to-a-given-number/ […]
… [Trackback]
[…] Find More Information here on that Topic: geeksforgeeks.org/shortest-root-to-leaf-path-sum-equal-to-a-given-number/ […]
… [Trackback]
[…] Find More on that Topic: geeksforgeeks.org/shortest-root-to-leaf-path-sum-equal-to-a-given-number/ […]
… [Trackback]
[…] Read More on to that Topic: geeksforgeeks.org/shortest-root-to-leaf-path-sum-equal-to-a-given-number/ […]
… [Trackback]
[…] Find More on on that Topic: geeksforgeeks.org/shortest-root-to-leaf-path-sum-equal-to-a-given-number/ […]