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Find all even sum paths in given Binary Search Tree

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  • Difficulty Level : Medium
  • Last Updated : 14 Jul, 2022
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Given a Binary search tree having N nodes, the task is to find all the paths starting at the root and ending at any leaf and having an even sum. 

Examples:

Input:

Img-Btree

Output:
Even sum Paths are:
1st) 1 -> 19 -> 4 -> 9 -> 7 = sum(40) 
2nd) 1 -> 19 -> 4 -> 9 -> 13 -> 18 -> 16 = sum(80)
3rd) 1 -> 19 -> 25 -> 35 = sum(68)
4th) 1 -> 19 -> 25 -> 23 = sum(80)
Explanation: When we start traversing form root node to the leaf node then we have different path are (1, 19, 4, 9, 7), (1, 19, 4, 9, 13, 18, 16), (1, 19, 25, 35), (1, 19, 25, 23) and (1, 19, 4, 2, 3). And after finding the sum of every path we found that all are even. Here we found one path as odd which are 1 -> 19 -> 4 -> 2 -> 3 = sum(29).

Input:    2
            /  \
         1     3
Output: No path
Explanation: There is no path which has even sum.

 

Approach: The problem can be solved using a stack, based on the following idea:

Traverse the whole tree and keep storing the path in the stack. When the leaf is reached check if the sum of the path is even or not.

Follow the steps mentioned below to implement the approach:

  • Take one stack (say path) to store the current path.
  • Recursively traverse the tree and in each iteration:
    • Store the sum of all the nodes along this path (say sum) and store that node data in path also.
    • Traverse for the left child.
    • Traverse for the right child.
    • If a leaf node is reached, check if the sum of the path is even or odd.
  • Return all such paths.

Below is the code to understand better:

C++




// C++ code to implement the approach
 
#include <bits/stdc++.h>
 
using namespace std;
 
int cnt;
 
// Structure of a tree node
class Node {
public:
    Node* left;
    int val;
    Node* right;
 
    Node(int val)
    {
        this->right = right;
        this->val = val;
        this->left = left;
    }
};
 
// Function to print path
void print(stack<int> pth)
{
    vector<int> v;
    cout << "[";
    while (!pth.empty()) {
        v.push_back(pth.top());
        pth.pop();
    }
    while (v.size() != 1) {
        cout << v.back() << ",";
        v.pop_back();
    }
    cout << v.back();
    cout << "]";
}
 
// Function to find the paths
void cntEvenPath(Node* root, int sum, stack<int>& path)
{
   
    // Add the value of node in sum
    sum += root->val;
 
    // Keep storing the node element
    // in the path
    path.push(root->val);
 
    // If node has no left and right child
    // and also the sum is even
    // then return the path
    if ((sum & 1) == 0
        && (root->left == NULL && root->right == NULL)) {
        cnt++;
        cout << "Sum Of ";
        print(path);
        cout << " = " << sum << "\n";
        return;
    }
 
    // Check left child
    if (root->left != NULL) {
        cntEvenPath(root->left, sum, path);
        path.pop();
    }
 
    // Check right child
    if (root->right != NULL) {
        cntEvenPath(root->right, sum, path);
        path.pop();
    }
}
 
// Driver code
int main()
{
    // Build a tree
    Node* root = new Node(1);
    root->right = new Node(19);
    root->right->right = new Node(25);
    root->right->left = new Node(4);
    root->right->left->left = new Node(2);
    root->right->left->right = new Node(9);
    root->right->right->left = new Node(23);
    root->right->right->right = new Node(35);
    root->right->left->left->right = new Node(3);
    root->right->left->right->left = new Node(7);
    root->right->left->right->right = new Node(13);
    root->right->left->right->right->right = new Node(18);
    root->right->left->right->right->right->left
        = new Node(16);
 
    cout << "\n";
 
    // Stack to store path
    stack<int> path;
    cntEvenPath(root, 0, path);
    if (cnt == 0)
        cout << "No path";
    return 0;
}
 
// This code is contributed by Rohit Pradhan


Java




// Java code to implement the approach
 
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
    static int count = 0;
 
    // Structure of a tree node
    static class Node {
        Node left;
        int val;
        Node right;
 
        Node(int val)
        {
            this.right = right;
            this.val = val;
            this.left = left;
        }
    }
 
    // Function to find the paths
    static void cntEvenPath(Node root, int sum,
                            Stack<Integer> path)
    {
        // Add the value of node in sum
        sum += root.val;
 
        // Keep storing the node element
        // in the path
        path.push(root.val);
 
        // If node has no left and right child
        // and also the sum is even
        // then return the path
        if ((sum & 1) == 0
            && (root.left == null
                && root.right == null)) {
            count++;
            System.out.println("Sum Of"
                               + path + " = "
                               + sum);
            return;
        }
 
        // Check left child
        if (root.left != null) {
            cntEvenPath(root.left, sum, path);
            path.pop();
        }
 
        // Check right child
        if (root.right != null) {
            cntEvenPath(root.right, sum, path);
            path.pop();
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        // Build a tree
        Node root = new Node(1);
        root.right = new Node(19);
        root.right.right = new Node(25);
        root.right.left = new Node(4);
        root.right.left.left = new Node(2);
        root.right.left.right = new Node(9);
        root.right.right.left = new Node(23);
        root.right.right.right
            = new Node(35);
        root.right.left.left.right
            = new Node(3);
        root.right.left.right.left
            = new Node(7);
        root.right.left.right.right
            = new Node(13);
        root.right.left.right.right.right
            = new Node(18);
        root.right.left.right.right.right.left
            = new Node(16);
 
        System.out.println();
 
        // Stack to store path
        Stack<Integer> path = new Stack<>();
        cntEvenPath(root, 0, path);
        if (count == 0)
            System.out.println("No path");
    }
}


Python3




# Python code to implement the approach
 
# Node to create the tree
class Node:
    def __init__(self, val):
        self.right = None
        self.val = val
        self.left = None
 
# Function to find the paths
def cntEvenPath(root, sum, path):
   
    # Add the value of node in sum
    sum += root.val
 
    # Keep storing the node element
    # in the path
    path.append(root.val)
 
    # If node has no left and right child
    # and also the sum is even
    # then return the path
    if ((sum & 1) == 0 and root.left == None and root.right == None):
        global count
        count += 1
        print("Sum Of", path, " = ", sum)
        return
 
    # Check left child
    if (root.left != None):
        cntEvenPath(root.left, sum, path)
        path.pop()
 
    # Check right child
    if (root.right != None):
        cntEvenPath(root.right, sum, path)
        path.pop()
 
count = 0
 
# Defining main function
def main():
    # Build a tree
    root = Node(1)
    root.right = Node(19)
    root.right.right = Node(25)
    root.right.left = Node(4)
    root.right.left.left = Node(2)
    root.right.left.right = Node(9)
    root.right.right.left = Node(23)
    root.right.right.right = Node(35)
    root.right.left.left.right = Node(3)
    root.right.left.right.left = Node(7)
    root.right.left.right.right = Node(13)
    root.right.left.right.right.right = Node(18)
    root.right.left.right.right.right.left = Node(16)
 
    # Stack to store path
    path = []
    cntEvenPath(root, 0, path)
    if (count == 0):
        print("No path")
 
 
if __name__ == "__main__":
    main()
 
# This code is contributed by jainlovely450


C#




// C# code to implement the approach
using System;
using System.Collections;
 
public class GFG {
 
  static int count = 0;
 
  // Structure of a tree node
  class Node {
    public Node left;
    public int val;
    public Node right;
 
    public Node(int val)
    {
      this.right = null;
      this.val = val;
      this.left = null;
    }
  }
 
  // Function to find the paths
  static void cntEvenPath(Node root, int sum, Stack path)
  {
    // Add the value of node in sum
    sum += root.val;
 
    // Keep storing the node element
    // in the path
    path.Push(root.val);
 
    // If node has no left and right child
    // and also the sum is even
    // then return the path
    if ((sum & 1) == 0
        && (root.left == null && root.right == null)) {
      count++;
      Console.Write("Sum Of [");
      Object[] arr = path.ToArray();
      for (int i = arr.Length - 1; i >= 0; i--) {
        if (i == 0) {
          Console.Write(arr[i]);
          break;
        }
        Console.Write(arr[i] + ",");
      }
      Console.WriteLine("] = " + sum);
      return;
    }
 
    // Check left child
    if (root.left != null) {
      cntEvenPath(root.left, sum, path);
      path.Pop();
    }
 
    // Check right child
    if (root.right != null) {
      cntEvenPath(root.right, sum, path);
      path.Pop();
    }
  }
 
  static public void Main()
  {
 
    // Build a tree
    Node root = new Node(1);
    root.right = new Node(19);
    root.right.right = new Node(25);
    root.right.left = new Node(4);
    root.right.left.left = new Node(2);
    root.right.left.right = new Node(9);
    root.right.right.left = new Node(23);
    root.right.right.right = new Node(35);
    root.right.left.left.right = new Node(3);
    root.right.left.right.left = new Node(7);
    root.right.left.right.right = new Node(13);
    root.right.left.right.right.right = new Node(18);
    root.right.left.right.right.right.left
      = new Node(16);
 
    Console.WriteLine();
 
    // Stack to store path
    Stack path = new Stack();
    cntEvenPath(root, 0, path);
    if (count == 0)
      Console.WriteLine("No path");
  }
}
 
// This code is contributed by lokesh(lokeshmvs21).


Output

Sum Of [1,19,4,9,7] = 40
Sum Of [1,19,4,9,13,18,16] = 80
Sum Of [1,19,25,23] = 68
Sum Of [1,19,25,35] = 80

Time Complexity: O(N)
Auxiliary Space: O(N)


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