Print all root-to-leaf paths with maximum count of even nodes

Given a Binary tree, the task is to print all possible root-to-leaf paths having a maximum number of even valued nodes.

Examples:

Input:

             2
            / \
           6   3
          / \   \
         4   7   11
        / \   \
       10  12   1

Output:
2 -> 6 -> 4 -> 10 
2 -> 6 -> 4 -> 12 
Explanation:
Count of even nodes on the path 2 -> 6 -> 4 -> 10 = 4 
Count of even nodes on the path 2 -> 6 -> 4 -> 12 = 4 
Count of even nodes on the path 2 -> 6 -> 7 -> 1 = 2 
Count of even nodes on the path 2 -> 3 -> 11 = 1 
Therefore, the paths consisting of maximum count of even nodes are {2 -> 6 -> 4 -> 10} and {2 -> 6 -> 4 -> 12 }. 

Input:

             5
            / \
           6   3
          / \   \
         4   9   2

Output: 5 -> 6 -> 4. 

Approach: The problem can be solved using Preorder Traversal. The idea is to traverse the given Binary Tree and find the maximum count of even nodes possible in a root-to-leaf path. Finally, traverse the tree and print all possible root-to-leaf paths having the maximum count of even nodes. Follow the steps below to solve the problem:

• Initialize two variables, say cntMaxEven and cntEven to store the maximum count of even nodes from the root to a leaf node and count of even nodes from root to a leaf node.
• Traverse the tree and check the following conditions: 
  • Check if current node is a leaf node and cntEven > cntMaxEven or not. If found to be true then update cntMaxEven = cntEven.
  • Otherwise, check if the current node is an even node or not. If found to be true, then Update cntEven += 1.
• Finally, traverse the tree and print all possible root-to-leaf paths having a count of even nodes equal to cntMaxEven.

C++

// C++ program to implement
// the above approach

#include <bits/stdc++.h>
using namespace std;

// Structure of the binary tree
struct Node {
    // Stores data
    int data;

    // Stores left child
    Node* left;

    // Stores right child
    Node* right;

    // Initialize a node
    // of binary tree
    Node(int val)
    {

        // Update data;
        data = val;
        left = right = NULL;
    }
};

// Function to find maximum count
// of even nodes in a path
int cntMaxEvenNodes(Node* root)
{

    // If the tree is an
    // empty binary tree.
    if (root == NULL) {
        return 0;
    }

    // Stores count of even nodes
    // in current subtree
    int cntEven = 0;

    // If root node is
    // an even node
    if (root->data % 2
        == 0) {

        // Update cntEven
        cntEven += 1;
    }

    // Stores count of even nodes
    // in left subtree.
    int X = cntMaxEvenNodes(
        root->left);

    // Stores count of even nodes
    // in right subtree.
    int Y = cntMaxEvenNodes(
        root->right);

    // cntEven
    cntEven += max(X, Y);

    return cntEven;
}

// Function to print paths having
// count of even nodes equal
// to maximum count of even nodes
void printPath(Node* root, int cntEven,
               int cntMaxEven,
               vector<int>& path)
{
    // If the tree is an
    // empty Binary Tree
    if (root == NULL) {
        return;
    }

    // If current node value is even
    if (root->data % 2 == 0) {
        path.push_back(
            root->data);
        cntEven += 1;
    }

    // If current node is
    // a leaf node
    if (root->left == NULL
        && root->right == NULL) {

        // If count of even nodes in
        // path is equal to cntMaxEven
        if (cntEven == cntMaxEven) {

            // Stores length of path
            int N = path.size();

            // Print path
            for (int i = 0; i < N - 1;
                 i++) {
                cout << path[i] << " -> ";
            }
            cout << path[N - 1] << endl;
        }
    }

    // Left subtree
    printPath(root->left, cntEven,
              cntMaxEven, path);

    // Right subtree
    printPath(root->right, cntEven,
              cntMaxEven, path);

    // If current node is even
    if (root->data % 2 == 0) {
        path.pop_back();

        // Update cntEven
        cntEven--;
    }
}

// Utility Function to print path
// from root to leaf node having
// maximum count of even nodes
void printMaxPath(Node* root)
{
    // Stores maximum count of even
    // nodes of a path in the tree
    int cntMaxEven;

    cntMaxEven
        = cntMaxEvenNodes(root);

    // Stores path of tree having
    // even nodes
    vector<int> path;

    printPath(root, 0, cntMaxEven,
              path);
}

// Driver code
int main()
{
    // Create tree.
    Node* root = NULL;
    root = new Node(2);
    root->left = new Node(6);
    root->right = new Node(3);
    root->left->left = new Node(4);
    root->left->right = new Node(7);
    root->right->right = new Node(11);
    root->left->left->left = new Node(10);
    root->left->left->right = new Node(12);
    root->left->right->right = new Node(1);

    printMaxPath(root);
}

Java

// Java program to implement
// the above approach
import java.util.*;

class GFG{

// Structure of the binary tree
static class Node
{

    // Stores data
    int data;

    // Stores left child
    Node left;

    // Stores right child
    Node right;

    // Initialize a node
    // of binary tree
    Node(int val)
    {

        // Update data;
        data = val;
        left = right = null;
    }
};

// Function to find maximum count
// of even nodes in a path
static int cntMaxEvenNodes(Node root)
{

    // If the tree is an
    // empty binary tree.
    if (root == null)
    {
        return 0;
    }

    // Stores count of even nodes
    // in current subtree
    int cntEven = 0;

    // If root node is
    // an even node
    if (root.data % 2 == 0)
    {

        // Update cntEven
        cntEven += 1;
    }

    // Stores count of even nodes
    // in left subtree.
    int X = cntMaxEvenNodes(root.left);

    // Stores count of even nodes
    // in right subtree.
    int Y = cntMaxEvenNodes(root.right);

    // cntEven
    cntEven += Math.max(X, Y);

    return cntEven;
}

// Function to print paths having
// count of even nodes equal
// to maximum count of even nodes
static void printPath(Node root, int cntEven,
                      int cntMaxEven,
                      Vector<Integer> path)
{

    // If the tree is an
    // empty Binary Tree
    if (root == null)
    {
        return;
    }

    // If current node value is even
    if (root.data % 2 == 0)
    {
        path.add(root.data);
        cntEven += 1;
    }

    // If current node is
    // a leaf node
    if (root.left == null &&
       root.right == null)
    {

        // If count of even nodes in
        // path is equal to cntMaxEven
        if (cntEven == cntMaxEven)
        {

            // Stores length of path
            int N = path.size();

            // Print path
            for(int i = 0; i < N - 1; i++)
            {
                System.out.print(path.get(i) + " -> ");
            }
            System.out.print(path.get(N - 1) + "\n");
        }
    }

    // Left subtree
    printPath(root.left, cntEven,
              cntMaxEven, path);

    // Right subtree
    printPath(root.right, cntEven,
              cntMaxEven, path);

    // If current node is even
    if (root.data % 2 == 0)
    {
        path.remove(path.size() - 1);

        // Update cntEven
        cntEven--;
    }
}

// Utility Function to print path
// from root to leaf node having
// maximum count of even nodes
static void printMaxPath(Node root)
{

    // Stores maximum count of even
    // nodes of a path in the tree
    int cntMaxEven;

    cntMaxEven = cntMaxEvenNodes(root);

    // Stores path of tree having
    // even nodes
    Vector<Integer> path = new Vector<>();

    printPath(root, 0, cntMaxEven,
              path);
}

// Driver code
public static void main(String[] args)
{

    // Create tree.
    Node root = null;
    root = new Node(2);
    root.left = new Node(6);
    root.right = new Node(3);
    root.left.left = new Node(4);
    root.left.right = new Node(7);
    root.right.right = new Node(11);
    root.left.left.left = new Node(10);
    root.left.left.right = new Node(12);
    root.left.right.right = new Node(1);

    printMaxPath(root);
}
}

// This code is contributed by Amit Katiyar

Python3

# Python3 program to implement
# the above approach

# Structure of the binary tree
class newNode:

    def __init__(self, val):

        self.data = val
        self.left = None
        self.right = None

path = []

# Function to find maximum count
# of even nodes in a path
def cntMaxEvenNodes(root):

    # If the tree is an
    # empty binary tree.
    if (root == None):
        return 0

    # Stores count of even nodes
    # in current subtree
    cntEven = 0

    # If root node is
    # an even node
    if (root.data % 2 == 0):

        # Update cntEven
        cntEven += 1

    # Stores count of even nodes
    # in left subtree.
    X = cntMaxEvenNodes(root.left)

    # Stores count of even nodes
    # in right subtree.
    Y = cntMaxEvenNodes(root.right)

    # cntEven
    cntEven += max(X, Y)

    return cntEven

# Function to print paths having
# count of even nodes equal
# to maximum count of even nodes
def printPath(root, cntEven, cntMaxEven):

    global path

    # If the tree is an
    # empty Binary Tree
    if (root == None):
        return

    # If current node value is even
    if (root.data % 2 == 0):
        path.append(root.data)
        cntEven += 1

    # If current node is
    # a leaf node
    if (root.left == None and
       root.right == None):

        # If count of even nodes in
        # path is equal to cntMaxEven
        if (cntEven == cntMaxEven):

            # Stores length of path
            N = len(path)

            # Print path
            for i in range(N - 1):
                print(path[i], end = "->")

            print(path[N - 1])

    # Left subtree
    printPath(root.left, cntEven, cntMaxEven)

    # Right subtree
    printPath(root.right, cntEven, cntMaxEven)

    # If current node is even
    if (root.data % 2 == 0):
        path.remove(path[len(path) - 1])

        # Update cntEven
        cntEven -= 1

# Utility Function to print path
# from root to leaf node having
# maximum count of even nodes
def printMaxPath(root):

    global path

    # Stores maximum count of even
    # nodes of a path in the tree
    cntMaxEven = cntMaxEvenNodes(root)

    printPath(root, 0, cntMaxEven)

# Driver code
if __name__ == '__main__':

    # Create tree.
    root = None
    root = newNode(2)
    root.left = newNode(6)
    root.right = newNode(3)
    root.left.left = newNode(4)
    root.left.right = newNode(7)
    root.right.right = newNode(11)
    root.left.left.left = newNode(10)
    root.left.left.right = newNode(12)
    root.left.right.right = newNode(1)

    printMaxPath(root)

# This code is contributed by bgangwar59

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{

// Structure of the binary tree
class Node
{
  // Stores data
  public int data;

  // Stores left child
  public Node left;

  // Stores right child
  public Node right;

  // Initialize a node
  // of binary tree
  public Node(int val)
  {
    // Update data;
    data = val;
    left = right = null;
  }
};

// Function to find maximum count
// of even nodes in a path
static int cntMaxEvenNodes(Node root)
{
  // If the tree is an
  // empty binary tree.
  if (root == null)
  {
    return 0;
  }

  // Stores count of even
  // nodes in current subtree
  int cntEven = 0;

  // If root node is
  // an even node
  if (root.data % 2 == 0)
  {
    // Update cntEven
    cntEven += 1;
  }

  // Stores count of even
  // nodes in left subtree.
  int X = cntMaxEvenNodes(root.left);

  // Stores count of even nodes
  // in right subtree.
  int Y = cntMaxEvenNodes(root.right);

  // cntEven
  cntEven += Math.Max(X, Y);

  return cntEven;
}

// Function to print paths having
// count of even nodes equal
// to maximum count of even nodes
static void printPath(Node root,
                      int cntEven,
                      int cntMaxEven,
                      List<int> path)
{
  // If the tree is an
  // empty Binary Tree
  if (root == null)
  {
    return;
  }

  // If current node value
  // is even
  if (root.data % 2 == 0)
  {
    path.Add(root.data);
    cntEven += 1;
  }

  // If current node is
  // a leaf node
  if (root.left == null &&
      root.right == null)
  {
    // If count of even nodes in
    // path is equal to cntMaxEven
    if (cntEven == cntMaxEven)
    {
      // Stores length of path
      int N = path.Count;

      // Print path
      for(int i = 0;
              i < N - 1; i++)
      {
        Console.Write(path[i] +
                      " -> ");
      }
      Console.Write(path[N - 1] +
                    "\n");
    }
  }

  // Left subtree
  printPath(root.left, cntEven,
            cntMaxEven, path);

  // Right subtree
  printPath(root.right, cntEven,
            cntMaxEven, path);

  // If current node is even
  if (root.data % 2 == 0)
  {
    path.RemoveAt(path.Count - 1);

    // Update cntEven
    cntEven--;
  }
}

// Utility Function to print path
// from root to leaf node having
// maximum count of even nodes
static void printMaxPath(Node root)
{
  // Stores maximum count of even
  // nodes of a path in the tree
  int cntMaxEven;

  cntMaxEven = cntMaxEvenNodes(root);

  // Stores path of tree having
  // even nodes
  List<int> path = new List<int>();

  printPath(root, 0,
            cntMaxEven, path);
}

// Driver code
public static void Main(String[] args)
{
  // Create tree.
  Node root = null;
  root = new Node(2);
  root.left = new Node(6);
  root.right = new Node(3);
  root.left.left = new Node(4);
  root.left.right = new Node(7);
  root.right.right = new Node(11);
  root.left.left.left = new Node(10);
  root.left.left.right = new Node(12);
  root.left.right.right = new Node(1);

  printMaxPath(root);
}
}

// This code is contributed by Princi Singh

Javascript

<script>

// Javascript program to implement
// the above approach

// Structure of the binary tree
class Node
{
    constructor(val)
    {
        this.left = null;
        this.right = null;
        this.data = val;
    }
}

// Function to find maximum count
// of even nodes in a path
function cntMaxEvenNodes(root)
{

    // If the tree is an
    // empty binary tree.
    if (root == null)
    {
        return 0;
    }

    // Stores count of even nodes
    // in current subtree
    let cntEven = 0;

    // If root node is
    // an even node
    if (root.data % 2 == 0)
    {

        // Update cntEven
        cntEven += 1;
    }

    // Stores count of even nodes
    // in left subtree.
    let X = cntMaxEvenNodes(root.left);

    // Stores count of even nodes
    // in right subtree.
    let Y = cntMaxEvenNodes(root.right);

    // cntEven
    cntEven += Math.max(X, Y);

    return cntEven;
}

// Function to print paths having
// count of even nodes equal
// to maximum count of even nodes
function printPath(root, cntEven,
                   cntMaxEven, path)
{

    // If the tree is an
    // empty Binary Tree
    if (root == null)
    {
        return;
    }

    // If current node value is even
    if (root.data % 2 == 0)
    {
        path.push(root.data);
        cntEven += 1;
    }

    // If current node is
    // a leaf node
    if (root.left == null &&
       root.right == null)
    {

        // If count of even nodes in
        // path is equal to cntMaxEven
        if (cntEven == cntMaxEven)
        {

            // Stores length of path
            let N = path.length;

            // Print path
            for(let i = 0; i < N - 1; i++)
            {
                document.write(path[i] + " -> ");
            }
            document.write(path[N - 1] + "</br>");
        }
    }

    // Left subtree
    printPath(root.left, cntEven,
              cntMaxEven, path);

    // Right subtree
    printPath(root.right, cntEven,
              cntMaxEven, path);

    // If current node is even
    if (root.data % 2 == 0)
    {
        path.pop();

        // Update cntEven
        cntEven--;
    }
}

// Utility Function to print path
// from root to leaf node having
// maximum count of even nodes
function printMaxPath(root)
{

    // Stores maximum count of even
    // nodes of a path in the tree
    let cntMaxEven;

    cntMaxEven = cntMaxEvenNodes(root);

    // Stores path of tree having
    // even nodes
    let path = [];

    printPath(root, 0, cntMaxEven, path);
}

// Driver code

// Create tree.
let root = null;
root = new Node(2);
root.left = new Node(6);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(7);
root.right.right = new Node(11);
root.left.left.left = new Node(10);
root.left.left.right = new Node(12);
root.left.right.right = new Node(1);

printMaxPath(root);

// This code is contributed by suresh07

</script>

Output:

2 -> 6 -> 4 -> 10
2 -> 6 -> 4 -> 12

Time Complexity: O(N), where N is the number of nodes in the binary tree. 
Auxiliary Space: O(N)