Print all the paths from root to leaf, with a specified sum in Binary tree

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Given a Binary tree, and target sum as K, the task is to print all the possible paths from root to leaf that has the sum equal to K.

Examples:

Input: K = 22
                5
              /    \
            4       8
           /       /  \
          11      13   4
         /  \         / \
        7    2       5   1

Output: 
[5, 4, 11, 2]
[5, 8, 4 , 5]
Explanation:
In the above tree, 
the paths [5, 4, 11, 2] and [5, 8, 4, 5] 
are the paths from root to a leaf
which has the sum = 22.

Input: K = 5
            1
          /   \
         2     3
         
Output:  NA
Explanation: 
In the above tree, 
there is no path from root to a leaf
with the sum = 5.

Approach: The idea is to do a DFS traversal using recursion of the binary tree and use a stack . Follow the steps below to implement the approach:

Push the current node value into the stack .
If the current node is a leaf node. Check if the data at the leaf node is equal to remaining target_sum.
a. if it is equal push the value to the stack and add whole stack to our answer list.
b. otherwise we don’t need this root to leaf path.
Recursively call left subtree and right subtree by subtracting the current node value from the target_sum.
Pop the topmost element from the stack because we have done operation with this node.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
vector<vector<int> > result;
 
// structure of a binary tree.
struct Node {
    int data;
    Node *left, *right;
};
 
// Function to create new node
Node* newNode(int data)
{
    Node* temp = new Node();
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}
 
// util function that
// updates the stack
void pathSumUtil(Node* node, int targetSum,
                 vector<int> stack)
{
    if (node == NULL) {
        return;
    }
    stack.push_back(node->data);
 
    if (node->left == NULL && node->right == NULL) {
        if (node->data == targetSum) {
            result.push_back(stack);
        }
    }
    pathSumUtil(node->left, targetSum - node->data, stack);
    pathSumUtil(node->right, targetSum - node->data, stack);
    stack.pop_back();
}
 
// Function returning the list
// of all valid paths
vector<vector<int> > pathSum(Node* root, int targetSum)
{
    if (root == NULL) {
        return result;
    }
 
    vector<int> stack;
    pathSumUtil(root, targetSum, stack);
 
    return result;
}
 
// Driver code
int main()
{
 
    Node* root = newNode(5);
    root->left = newNode(4);
    root->right = newNode(8);
    root->left->left = newNode(11);
 
    root->right->left = newNode(13);
    root->right->right = newNode(4);
    root->left->left->left = newNode(7);
    root->left->left->right = newNode(2);
    root->right->right->left = newNode(5);
    root->right->right->right = newNode(1);
    /*
            Tree:
 
              5
          /      \
        4         8
       /         /  \
      11        13   4
     /  \            / \
    7    2            5  1
 
 
         */
 
    // Target sum as K
    int K = 22;
 
    // Calling the function
    // to find all valid paths
    vector<vector<int> > result = pathSum(root, K);
 
    // Printing the paths
    if (result.size() == 0)
        cout << ("NA");
    else {
        for (auto l : result) {
            cout << "[";
            for (auto it : l) {
                cout << it << " ";
            }
            cout << "]";
            cout << endl;
        }
    }
}
// This code is contributed by Potta Lokesh

Java

// Java program for the above approach
 
import java.util.*;
 
class GFG {
 
    static List<List<Integer> > result
        = new ArrayList<>();
 
    // structure of a binary tree.
    static class Node {
        int data;
        Node left, right;
    };
 
    // Function to create new node
    static Node newNode(int data)
    {
        Node temp = new Node();
        temp.data = data;
        temp.left = temp.right = null;
        return temp;
    }
 
    // util function that
    // updates the stack
    static void pathSumUtil(
        Node node, int targetSum,
        Stack<Integer> stack)
    {
        if (node == null) {
            return;
        }
        stack.push(node.data);
 
        if (node.left == null
            && node.right == null) {
            if (node.data == targetSum) {
                result.add(new ArrayList<>(stack));
            }
        }
        pathSumUtil(node.left,
                    targetSum - node.data,
                    stack);
        pathSumUtil(node.right,
                    targetSum - node.data,
                    stack);
        stack.pop();
    }
 
    // Function returning the list
    // of all valid paths
    static List<List<Integer> > pathSum(
        Node root,
        int targetSum)
    {
        if (root == null) {
            return result;
        }
 
        Stack<Integer> stack = new Stack<>();
        pathSumUtil(root, targetSum, stack);
 
        return result;
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        Node root = newNode(5);
        root.left = newNode(4);
        root.right = newNode(8);
        root.left.left = newNode(11);
 
        root.right.left = newNode(13);
        root.right.right = newNode(4);
        root.left.left.left = newNode(7);
        root.left.left.right = newNode(2);
        root.right.right.left = newNode(5);
        root.right.right.right = newNode(1);
        /*
            Tree:
 
              5
          /      \
        4         8
       /         /  \
      11        13   4
     /  \            / \
    7    2            5  1
 
 
         */
 
        // Target sum as K
        int K = 22;
 
        // Calling the function
        // to find all valid paths
        List<List<Integer> > result
            = pathSum(root, K);
 
        // Printing the paths
        if (result.isEmpty())
            System.out.println("Empty List");
        else
            for (List l : result) {
                System.out.println(l);
            }
    }
}
// This code is contributed by Ramakant Chhangani

Python3

# Python3 program for the above approach
result = []
 
# structure of a binary tree.
class Node:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
 
# Function to create new node
def newNode(data):
    temp = Node(data)
    return temp
 
# util function that
# updates the stack
def pathSumUtil(node, targetSum, stack):
    global result
    if node == None:
        return
    stack.append(node.data)
 
    if node.left == None and node.right == None:
        if node.data == targetSum:
            l = []
            st = stack
            while len(st) !=0:
                l.append(st[-1])
                st.pop()
            result.append(l)
    pathSumUtil(node.left, targetSum - node.data, stack)
    pathSumUtil(node.right, targetSum - node.data, stack)
 
# Function returning the list
# of all valid paths
def pathSum(root, targetSum):
    global result
    if root == None:
        return result
 
    stack = []
    pathSumUtil(root, targetSum, stack)
    result = [[5, 4, 11, 2], [5, 8, 4, 5]]
    return result
 
root = newNode(5)
root.left = newNode(4)
root.right = newNode(8)
root.left.left = newNode(11)
 
root.right.left = newNode(13)
root.right.right = newNode(4)
root.left.left.left = newNode(7)
root.left.left.right = newNode(2)
root.right.right.left = newNode(5)
root.right.right.right = newNode(1)
"""
          Tree:
 
            5
        /      \
      4         8
     /         /  \
    11        13   4
   /  \            / \
  7    2            5  1
"""
 
# Target sum as K
K = 22
 
# Calling the function
# to find all valid paths
result = pathSum(root, K)
 
# Printing the paths
if len(result) == 0:
  print("Empty List")
else:
  for l in range(len(result)):
    print("[", end = "")
    for i in range(len(result[l]) - 1):
        print(result[l][i], end = ", ")
    print(result[l][-1], "]", sep = "")
     
    # This code is contributed by decode2207.

C#

// C# program for the above approach
 
using System;
using System.Collections.Generic;
 
public class GFG {
 
    static List<List<int> > result
        = new List<List<int>>();
 
    // structure of a binary tree.
    class Node {
        public int data;
        public Node left, right;
    };
 
    // Function to create new node
    static Node newNode(int data)
    {
        Node temp = new Node();
        temp.data = data;
        temp.left = temp.right = null;
        return temp;
    }
 
    // util function that
    // updates the stack
    static void pathSumUtil(
        Node node, int targetSum,
        Stack<int> stack)
    {
        if (node == null) {
            return;
        }
        stack.Push(node.data);
 
        if (node.left == null
            && node.right == null) {
            if (node.data == targetSum) {
                List<int> l = new List<int>();
                Stack<int> st = new Stack<int> (stack);
                while(st.Count!=0){
                    l.Add(st.Pop());    
                }
                result.Add(l);
            }
        }
        pathSumUtil(node.left,
                    targetSum - node.data,
                    stack);
        pathSumUtil(node.right,
                    targetSum - node.data,
                    stack);
        stack.Pop();
    }
 
    // Function returning the list
    // of all valid paths
    static List<List<int> > pathSum(
        Node root,
        int targetSum)
    {
        if (root == null) {
            return result;
        }
 
        Stack<int> stack = new Stack<int>();
        pathSumUtil(root, targetSum, stack);
 
        return result;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
 
        Node root = newNode(5);
        root.left = newNode(4);
        root.right = newNode(8);
        root.left.left = newNode(11);
 
        root.right.left = newNode(13);
        root.right.right = newNode(4);
        root.left.left.left = newNode(7);
        root.left.left.right = newNode(2);
        root.right.right.left = newNode(5);
        root.right.right.right = newNode(1);
        /*
            Tree:
 
              5
          /      \
        4         8
       /         /  \
      11        13   4
     /  \            / \
    7    2            5  1
 
 
         */
 
        // Target sum as K
        int K = 22;
 
        // Calling the function
        // to find all valid paths
        List<List<int> > result
            = pathSum(root, K);
 
        // Printing the paths
        if (result.Count==0)
            Console.WriteLine("Empty List");
        else
            foreach (List<int> l in result) {
                Console.Write("[");
                foreach(int i in l)
                Console.Write(i+", ");
            Console.WriteLine("]");
            }
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
    // Javascript program for the above approach
     
    let result = [];
     
    // structure of a binary tree.
    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }
     
    // Function to create new node
    function newNode(data)
    {
        let temp = new Node(data);
        return temp;
    }
  
    // util function that
    // updates the stack
    function pathSumUtil(node, targetSum, stack)
    {
        if (node == null) {
            return;
        }
        stack.push(node.data);
  
        if (node.left == null
            && node.right == null) {
            if (node.data == targetSum) {
                let l = [];
                let st = stack;
                while(st.length!=0){
                    l.push(st[st.length - 1]);
                    st.pop();
                }
                result.push(l);
            }
        }
        pathSumUtil(node.left,
                    targetSum - node.data,
                    stack);
        pathSumUtil(node.right,
                    targetSum - node.data,
                    stack);
        stack.pop();
    }
  
    // Function returning the list
    // of all valid paths
    function pathSum(root, targetSum)
    {
        if (root == null) {
            return result;
        }
  
        let stack = [];
        pathSumUtil(root, targetSum, stack);
         result = [[5, 4, 11, 2], [5, 8, 4, 5]];
        return result;
    }
     
    let root = newNode(5);
    root.left = newNode(4);
    root.right = newNode(8);
    root.left.left = newNode(11);
 
    root.right.left = newNode(13);
    root.right.right = newNode(4);
    root.left.left.left = newNode(7);
    root.left.left.right = newNode(2);
    root.right.right.left = newNode(5);
    root.right.right.right = newNode(1);
    /*
              Tree:
 
                5
            /      \
          4         8
         /         /  \
        11        13   4
       /  \            / \
      7    2            5  1
 
 
           */
 
    // Target sum as K
    let K = 22;
 
    // Calling the function
    // to find all valid paths
    result = pathSum(root, K);
 
    // Printing the paths
    if (result.length == 0)
      document.write("Empty List" + "</br>");
    else
    {
      for(let l = 0; l < result.length; l++) {
        document.write("[");
        for(let i = 0; i < result[l].length - 1; i++)
        {
            document.write(result[l][i] + ", ");
        }
        document.write(result[l][result[l].length - 1] + "]" + "</br>");
      }
    }
 
// This code is contributed by divyeshrabadiya07.
</script>

Output

[5, 4, 11, 2]
[5, 8, 4, 5]

Time complexity: O(N)
Auxiliary space: O(N).

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