Minimize changes to convert into Tree with root 1, even left children and odd right children

Given, a binary tree, the task is to convert this tree using minimum number of increment-decrement operations into a tree which satisfies the following conditions:

• The root node is always 1.
• Every left child of a node is even.
• Every right child of a node is odd. 

Return and print the minimum number of increment-decrement operations required at the end. 

Examples:

Input: 

Output: 3
Explanation: 
Since root is already 1, no change is needed at root
Left child of root node is 2, so no change is needed here.
Right child of root node is 2, so change it to 3, making a change of 1.
Left child of node 2 is 5, so change it to 4, making a change of 1.
Left child of node 3 is 6, so no change is needed here.
Right child of node 3 is 8, so change it to 9, making a change of 1.
Hence total changes needed is 3.

Input:

Output: 0
Explanation: The given tree already satisfies the given conditions.

Approach: The idea is to change every node which does not satisfies the condition based on below observation:

• If the root node is not 1, we can keep decrementing it by 1 till it becomes 1. So root-1 operations needed here.
• If the current node is a left child, and it is odd, we can simply make it even by incrementing or decrementing by 1. So 1 operation is needed here.
• If the current node is a right child, and it is even, we can simply make it odd by incrementing or decrementing by 1. So 1 operation is needed here.

Therefore, simply traverse the given Tree and:

• Add root-1 operations to the answer if the root is not 1.
• Add the count of left child which are odd, to the answer
• Also add the count of right child which are even, to the answer.

Based on above idea, we can do a DFS traversal as per below steps:

• Traverse left and right child recursively.
  • Check if visited node’s left value is not null and node’s left child value is odd
    • Increment answer by 1
  • Check if visited node’s right value is not null and node’s right child value is even 
    • Increment answer by 1
• Again Recursively call for left and right node till whole tree is traversed.
• Also check if root is not equal to 1. 
  • If true, add root_value – 1 to the answer.
• Return the answer at the end.

Below is the implementation of the above approach:

C++

// C++ code for the above approach

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

// To store the final changes needed
int count_of_changes;

struct Node {
    int value;
    struct Node *left, *right;
};

// Utility function to create new
// tree node
Node* newNode(int data)
{
    Node* temp = new Node;
    temp->value = data;
    temp->left = temp->right = NULL;
    return temp;
}

// DFS function to convert
// binary tree to proper tree
void dfs(Node* root)
{

    // Check if given root is NULL
    // base case
    if (root == NULL)
        return;

    // Check if visited node's
    // left value is not null and
    // node's left child value is odd
    // decrement its value by 1
    if (root->left
        && root->left->value % 2 == 1) {
        root->left->value -= 1;
        count_of_changes++;
    }

    // Check if visited node's
    // value is not null and node's
    // right child value is even,
    // increment its value by 1
    if (root->right
        && root->right->value % 2 == 0) {
        root->right->value += 1;
        count_of_changes++;
    }

    // Recursive call for left node
    dfs(root->left);

    // Recursive call for right node
    dfs(root->right);
}

// Function to find
// the min changes needed
int minCount(Node* root)
{

    // Initial value for
    // final changes needed
    count_of_changes = 0;

    // Base case to check
    // if root is NULL
    if (root == NULL)
        return count_of_changes;

    if (root->value != 1) {

        // Add root_value - 1 to the ans
        count_of_changes += root->value - 1;

        // Set root->Value to 1
        root->value = 1;
    }

    // DFS Function call
    dfs(root);

    // Return the final count
    return count_of_changes;
}

// Driver Code
int main()
{

    // Taking input
    Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(2);
    root->left->left = newNode(5);
    root->right->left = newNode(6);
    root->right->right = newNode(8);

    // Function call
    cout << minCount(root) << endl;

    return 0;
}

Java

// JAVA code for the above approach
import java.util.*;
class GFG
{

  // To store the final changes needed
  static int count_of_changes;

  public static class Node {
    int value;
    Node left, right;
  }

  // Utility function to create new
  // tree node

  public static Node newNode(int data)
  {
    Node temp = new Node();
    temp.value = data;
    temp.left = temp.right = null;
    return temp;
  }
  // DFS function to convert
  // binary tree to proper tree
  public static void dfs(Node root)
  {

    // Check if given root is null
    // base case
    if (root == null)
      return;

    // Check if visited node's
    // left value is not null and
    // node's left child value is odd
    // decrement its value by 1
    if (root.left != null && root.left.value % 2 == 1) {
      root.left.value -= 1;
      count_of_changes++;
    }

    // Check if visited node's
    // value is not null and node's
    // right child value is even,
    // increment its value by 1
    if (root.right != null && root.right.value % 2 == 0) {
      root.right.value += 1;
      count_of_changes++;
    }

    // Recursive call for left node
    dfs(root.left);

    // Recursive call for right node
    dfs(root.right);
  }

  // Function to find
  // the min changes needed
  public static int minCount(Node root)
  {

    // Initial value for
    // final changes needed
    count_of_changes = 0;

    // Base case to check
    // if root is null
    if (root == null)
      return count_of_changes;

    if (root.value != 1) {

      // Add root_value - 1 to the ans
      count_of_changes += root.value - 1;

      // Set root->Value to 1
      root.value = 1;
    }

    // DFS Function call
    dfs(root);

    // Return the final count
    return count_of_changes;
  }


  // Driver code
  public static void main(String[] args)
  {
    // Taking input
    Node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(2);
    root.left.left = newNode(5);
    root.right.left = newNode(6);
    root.right.right = newNode(8);

    // Function call
    System.out.println(minCount(root));

  }
}

// This code is contributed by jana_sayantan.

Python3

# Python code for the above approach

# To store the final changes needed
count_of_changes = 0

class Node:
    def __init__(self,data = 0,left = None,right = None):
        self.data = data
        self.left = left
        self.right = right

# Utility function to create new
# tree node
def newNode(data):

    temp = Node()
    temp.value = data
    temp.left = temp.right = None
    return temp


# DFS function to convert
# binary tree to proper tree
def dfs(root):

    global count_of_changes

    # Check if given root is None
    # base case
    if (root == None):
        return

    # Check if visited node's
    # left value is not None and
    # node's left child value is odd
    # decrement its value by 1
    if (root.left and root.left.value % 2 == 1) :
        root.left.value -= 1
        count_of_changes += 1

    # Check if visited node's
    # value is not None and node's
    # right child value is even,
    # increment its value by 1
    if (root.right
        and root.right.value % 2 == 0):
        root.right.value += 1
        count_of_changes += 1

    # Recursive call for left node
    dfs(root.left)

    # Recursive call for right node
    dfs(root.right)

# Function to find
# the min changes needed
def minCount(root):

    # Initial value for
    # final changes needed
    global count_of_changes
    count_of_changes = 0

    # Base case to check
    # if root is None
    if (root == None):
        return count_of_changes

    if (root.value != 1):

        # Add root_value - 1 to the ans
        count_of_changes += root.value - 1

        # Set root.Value to 1
        root.value = 1

    # DFS Function call
    dfs(root)

    # Return the final count
    return count_of_changes

# Driver Code

# Taking input
root = newNode(1)
root.left = newNode(2)
root.right = newNode(2)
root.left.left = newNode(5)
root.right.left = newNode(6)
root.right.right = newNode(8)

# Function call
print(minCount(root))

# This code is contributed by shinjanpatra

C#

// C# program to implement above approach
using System;
using System.Collections;
using System.Collections.Generic;

class GFG
{

  // To store the final changes needed
  static int count_of_changes;

  public class Node {
    public int value;
    public Node left, right;
  }

  // Utility function to create new
  // tree node

  public static Node newNode(int data)
  {
    Node temp = new Node();
    temp.value = data;
    temp.left = temp.right = null;
    return temp;
  }

  // DFS function to convert
  // binary tree to proper tree
  public static void dfs(Node root)
  {

    // Check if given root is null
    // base case
    if (root == null)
      return;

    // Check if visited node's
    // left value is not null and
    // node's left child value is odd
    // decrement its value by 1
    if (root.left != null && root.left.value % 2 == 1) {
      root.left.value-=1;
      count_of_changes++;
    }

    // Check if visited node's
    // value is not null and node's
    // right child value is even,
    // increment its value by 1
    if (root.right != null && root.right.value % 2 == 0) {
      root.right.value += 1;
      count_of_changes++;
    }

    // Recursive call for left node
    dfs(root.left);

    // Recursive call for right node
    dfs(root.right);
  }

  // Function to find
  // the min changes needed
  public static int minCount(Node root)
  {

    // Initial value for
    // final changes needed
    count_of_changes = 0;

    // Base case to check
    // if root is null
    if (root == null)
      return count_of_changes;

    if (root.value != 1) {

      // Add root_value - 1 to the ans
      count_of_changes += root.value - 1;

      // Set root->Value to 1
      root.value = 1;
    }

    // DFS Function call
    dfs(root);

    // Return the final count
    return count_of_changes;
  }


  public static void Main(string[] args){

    // Taking input
    Node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(2);
    root.left.left = newNode(5);
    root.right.left = newNode(6);
    root.right.right = newNode(8);

    // Function call
    Console.WriteLine(minCount(root));

  }
}

// This code is contributed by entertain2022.

Javascript

<script>

// JavaScript code to implement the above approach

// To store the final changes needed
let count_of_changes = 0

class Node{
    constructor(data = 0,left = null,right = null){
        this.data = data
        this.left = left
        this.right = right
    }
}

// Utility function to create new
// tree node
function newNode(data){

    let temp = new Node()
    temp.value = data
    temp.left = temp.right = null
    return temp
}

// DFS function to convert
// binary tree to proper tree
function dfs(root){

    // Check if given root is null
    // base case
    if (root == null)
        return

    // Check if visited node's
    // left value is not null and
    // node's left child value is odd
    // decrement its value by 1
    if (root.left && root.left.value % 2 == 1){
        root.left.value -= 1
        count_of_changes += 1
    }

    // Check if visited node's
    // value is not null and node's
    // right child value is even,
    // increment its value by 1
    if (root.right
        && root.right.value % 2 == 0){
        root.right.value += 1
        count_of_changes += 1
    }

    // Recursive call for left node
    dfs(root.left)

    // Recursive call for right node
    dfs(root.right)
}

// Function to find
// the min changes needed
function minCount(root){

    // Initial value for
    // final changes needed
    count_of_changes = 0

    // Base case to check
    // if root is null
    if (root == null)
        return count_of_changes

    if (root.value != 1){

        // Add root_value - 1 to the ans
        count_of_changes += root.value - 1

        // Set root.Value to 1
        root.value = 1
    }

    // DFS Function call
    dfs(root)

    // Return the final count
    return count_of_changes
}

// Driver Code

// Taking input
let root = newNode(1)
root.left = newNode(2)
root.right = newNode(2)
root.left.left = newNode(5)
root.right.left = newNode(6)
root.right.right = newNode(8)

// Function call
document.write(minCount(root),"</br>")

// This code is contributed by shinjanpatra

</script>

Output

3

Time Complexity: O(V), where V is the count of vertices in given Tree
Auxiliary Space: O(H), which is the size of the stack for function calls, where H is the height of the tree.