Minimum sum path between two leaves of a binary tree
Last Updated :
06 Aug, 2021
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 node
struct Node {
int data;
struct Node* left;
struct Node* right;
};
// Utility function to allocate memory
// for a new node
struct 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_MIN
int 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 leaves
int minPathSum(struct Node* root)
{
int result = INT_MAX;
minPathSumUtil(root, result);
return result;
}
// Driver code
int 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 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 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# 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
<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>
Time Complexity: O(N)
Auxiliary Space: O(N)
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