Given a binary tree in which each node element contains a number. The task is to find the minimum possible sum from one leaf node to another.
If one side of root is empty, then function should return minus infinite.
Examples:
Input :
4
/ \
5 -6
/ \ / \
2 -3 1 8
Output : 1
The minimum sum path between two leaf nodes is:
-3 -> 5 -> 4 -> -6 -> 1
Input :
3
/ \
2 4
/ \
-5 1
Output : -2
Approach: The idea is to maintain two values in recursive calls:
- Minimum root to leaf path sum for the subtree rooted under current node.
- The minimum path sum between leaves.
For every visited node X, we find the minimum root to leaf sum in left and right sub trees of X. We add the two values with X’s data, and compare the sum with the current minimum path sum.
Below is the implementation of the above approach:
C++
// C++ program to find minimum path sum// between two leaves of a binary tree#include <bits/stdc++.h>using namespace std;// A binary tree nodestruct Node { int data; struct Node* left; struct Node* right;};// Utility function to allocate memory// for a new nodestruct Node* newNode(int data){ struct Node* node = new (struct Node); node->data = data; node->left = node->right = NULL; return (node);}// A utility function to find the minimum sum between// any two leaves. This function calculates two values:// 1. Minimum path sum between two leaves which is stored// in result and,// 2. The minimum root to leaf path sum which is returned.// If one side of root is empty, then it returns INT_MINint minPathSumUtil(struct Node* root, int& result){ // Base cases if (root == NULL) return 0; if (root->left == NULL && root->right == NULL) return root->data; // Find minimum sum in left and right sub tree. Also // find minimum root to leaf sums in left and right // sub trees and store them in ls and rs int ls = minPathSumUtil(root->left, result); int rs = minPathSumUtil(root->right, result); // If both left and right children exist if (root->left && root->right) { // Update result if needed result = min(result, ls + rs + root->data); // Return minimum possible value for root being // on one side return min(ls + root->data, rs + root->data); } // If any of the two children is empty, return // root sum for root being on one side if (root->left == NULL) return rs + root->data; else return ls + root->data;}// Function to return the minimum// sum path between two leavesint minPathSum(struct Node* root){ int result = INT_MAX; minPathSumUtil(root, result); return result;}// Driver codeint main(){ struct Node* root = newNode(4); root->left = newNode(5); root->right = newNode(-6); root->left->left = newNode(2); root->left->right = newNode(-3); root->right->left = newNode(1); root->right->right = newNode(8); cout << minPathSum(root); return 0;} |
Java
// Java program to find minimum path sum // between two leaves of a binary tree class GFG{ // A binary tree node static class Node { int data; Node left; Node right; }; // Utility function to allocate memory // for a new node static Node newNode(int data) { Node node = new Node(); node.data = data; node.left = node.right = null; return (node); } static int result;// A utility function to find the minimum sum between // any two leaves. This function calculates two values: // 1. Minimum path sum between two leaves which is stored // in result and, // 2. The minimum root to leaf path sum which is returned. // If one side of root is empty, then it returns INT_MIN static int minPathSumUtil( Node root) { // Base cases if (root == null) return 0; if (root.left == null && root.right == null) return root.data; // Find minimum sum in left and right sub tree. Also // find minimum root to leaf sums in left and right // sub trees and store them in ls and rs int ls = minPathSumUtil(root.left); int rs = minPathSumUtil(root.right); // If both left and right children exist if (root.left != null && root.right != null) { // Update result if needed result = Math.min(result, ls + rs + root.data); // Return minimum possible value for root being // on one side return Math.min(ls + root.data, rs + root.data); } // If any of the two children is empty, return // root sum for root being on one side if (root.left == null) return rs + root.data; else return ls + root.data; } // Function to return the minimum // sum path between two leaves static int minPathSum( Node root) { result = Integer.MAX_VALUE; minPathSumUtil(root); return result; } // Driver code public static void main(String args[]){ Node root = newNode(4); root.left = newNode(5); root.right = newNode(-6); root.left.left = newNode(2); root.left.right = newNode(-3); root.right.left = newNode(1); root.right.right = newNode(8); System.out.print(minPathSum(root)); }} // This code is contributed by Arnab Kundu |
Python3
# Python3 program to find minimum path sum # between two leaves of a binary tree # Tree node class Node: def __init__(self, data): self.data = data self.left = None self.right = None# Utility function to allocate memory # for a new node def newNode( data): node = Node(0) node.data = data node.left = node.right = None return (node) result = -1# A utility function to find the # minimum sum between any two leaves.# This function calculates two values: # 1. Minimum path sum between two leaves # which is stored in result and, # 2. The minimum root to leaf path sum # which is returned. # If one side of root is empty, # then it returns INT_MIN def minPathSumUtil(root) : global result # Base cases if (root == None): return 0 if (root.left == None and root.right == None) : return root.data # Find minimum sum in left and right sub tree. # Also find minimum root to leaf sums in # left and right sub trees and store them # in ls and rs ls = minPathSumUtil(root.left) rs = minPathSumUtil(root.right) # If both left and right children exist if (root.left != None and root.right != None) : # Update result if needed result = min(result, ls + rs + root.data) # Return minimum possible value for # root being on one side return min(ls + root.data, rs + root.data) # If any of the two children is empty, # return root sum for root being on one side if (root.left == None) : return rs + root.data else: return ls + root.data # Function to return the minimum # sum path between two leaves def minPathSum( root): global result result = 9999999 minPathSumUtil(root) return result # Driver code root = newNode(4) root.left = newNode(5) root.right = newNode(-6) root.left.left = newNode(2) root.left.right = newNode(-3) root.right.left = newNode(1) root.right.right = newNode(8) print(minPathSum(root)) # This code is contributed # by Arnab Kundu |
C#
// C# program to find minimum path sum // between two leaves of a binary tree using System; class GFG{ // A binary tree node public class Node { public int data; public Node left; public Node right; }; // Utility function to allocate memory // for a new node static Node newNode(int data) { Node node = new Node(); node.data = data; node.left = node.right = null; return (node); } static int result;// A utility function to find the minimum sum between // any two leaves. This function calculates two values: // 1. Minimum path sum between two leaves which is stored // in result and, // 2. The minimum root to leaf path sum which is returned. // If one side of root is empty, then it returns INT_MIN static int minPathSumUtil( Node root) { // Base cases if (root == null) return 0; if (root.left == null && root.right == null) return root.data; // Find minimum sum in left and right sub tree. Also // find minimum root to leaf sums in left and right // sub trees and store them in ls and rs int ls = minPathSumUtil(root.left); int rs = minPathSumUtil(root.right); // If both left and right children exist if (root.left != null && root.right != null) { // Update result if needed result = Math.Min(result, ls + rs + root.data); // Return minimum possible value for root being // on one side return Math.Min(ls + root.data, rs + root.data); } // If any of the two children is empty, return // root sum for root being on one side if (root.left == null) return rs + root.data; else return ls + root.data; } // Function to return the minimum // sum path between two leaves static int minPathSum( Node root) { result = int.MaxValue; minPathSumUtil(root); return result; } // Driver code public static void Main(String []args){ Node root = newNode(4); root.left = newNode(5); root.right = newNode(-6); root.left.left = newNode(2); root.left.right = newNode(-3); root.right.left = newNode(1); root.right.right = newNode(8); Console.Write(minPathSum(root)); }}// This code has been contributed by 29AjayKumar |
Javascript
<script> // JavaScript program to find minimum path sum // between two leaves of a binary tree // Structure of binary tree class Node { constructor(data) { this.left = null; this.right = null; this.data = data; } } // Utility function to allocate memory // for a new node function newNode(data) { let node = new Node(data); return (node); } let result; // A utility function to find the minimum sum between // any two leaves. This function calculates two values: // 1. Minimum path sum between two leaves which is stored // in result and, // 2. The minimum root to leaf path sum which is returned. // If one side of root is empty, then it returns INT_MIN function minPathSumUtil(root) { // Base cases if (root == null) return 0; if (root.left == null && root.right == null) return root.data; // Find minimum sum in left and right sub tree. Also // find minimum root to leaf sums in left and right // sub trees and store them in ls and rs let ls = minPathSumUtil(root.left); let rs = minPathSumUtil(root.right); // If both left and right children exist if (root.left != null && root.right != null) { // Update result if needed result = Math.min(result, ls + rs + root.data); // Return minimum possible value for root being // on one side return Math.min(ls + root.data, rs + root.data); } // If any of the two children is empty, return // root sum for root being on one side if (root.left == null) return rs + root.data; else return ls + root.data; } // Function to return the minimum // sum path between two leaves function minPathSum(root) { result = Number.MAX_VALUE; minPathSumUtil(root); return result; } let root = newNode(4); root.left = newNode(5); root.right = newNode(-6); root.left.left = newNode(2); root.left.right = newNode(-3); root.right.left = newNode(1); root.right.right = newNode(8); document.write(minPathSum(root)); </script> |
Output:
1
Time Complexity: O(N)
Auxiliary Space: O(N)
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