Min-Max Product Tree of a given Binary Tree
Last Updated :
15 Jul, 2025
Given a Binary Tree, the task is to convert the given Binary tree into Min-Max Product Tree and print the level-order sequence of the modified tree.
Min-Max Product Tree: A Min-Max Product Tree contains the product of the minimum and maximum values of its left and right subtrees at each node.
Note: For any node having only a single child, the value of that child node will be considered as both the minimum and the maximum.
Examples:
Input:
1
/ \
12 11
/ / \
3 4 13
\ /
15 5
Output:
45 9 60 3 225 25 15 5
Explanation:
Min-Max Product Tree:
45 (3 * 15)
/ \
( 3 * 3)9 60 (4 * 15)
/ / \
3 225 25 (5* 5)
\ /
15 5
Input:
5
/ \
21 77
/ \ \
61 16 16
\ /
10 3
/
23
Output:
231 610 48 61 230 9 529 3 23
Approach:
The idea is to use the post order traversal to traverse the left and right subtree recursively and extract the minimum and maximum. Store the product of minimum and maximum for every node in the Min-Max Product tree. Once, computed, compare the current node value with the current minimum and maximum and modify accordingly. Return the new minimum and maximum for the node at a higher level. Repeat this process for all nodes and finally, print the level order traversal of the Min-Max Product tree.
Below is the implementation of the above approach:
C++
// C++ implementation of
// the above approach
#include <bits/stdc++.h>
using namespace std;
// A Tree node
struct Node {
int data;
struct Node *left, *right;
};
// Utility function to create
// a new node
Node* newNode(int key)
{
Node* temp = new Node;
temp->data = key;
temp->left = temp->right = NULL;
return (temp);
}
// A minMax Structure for storing
// minimum and maximum value
struct minMax {
int min;
int max;
};
// Function to return min value
int min(minMax* a, minMax* b)
{
if (a != NULL
&& b != NULL) {
return a->min < b->min
? a->min
: b->min;
}
return a == NULL
? b->min
: a->min;
}
// Function to return max value
int max(minMax* a, minMax* b)
{
if (a != NULL
&& b != NULL) {
return a->max > b->max
? a->max
: b->max;
}
return a == NULL
? b->max
: a->max;
}
// Function to return min value
int min(int x, int y)
{
return x < y ? x : y;
}
// Function to return max value
int max(int x, int y)
{
return x > y ? x : y;
}
// Utility function to create
// min-max product tree
minMax* minMaxTreeUtil(
Node* root)
{
// Base condition
if (root == NULL) {
return NULL;
}
minMax* var = new minMax;
// Condition to check if
// current node is leaf node
if (root->left == NULL
&& root->right == NULL) {
var->min = root->data;
var->max = root->data;
return var;
}
// Left recursive call
minMax* left
= minMaxTreeUtil(root->left);
// Right recursive call
minMax* right
= minMaxTreeUtil(root->right);
// Store Min
var->min = min(left, right);
// Store Max
var->max = max(left, right);
// Store current node data
int currData = root->data;
// Assign product of minimum
// and maximum value
root->data = var->min * var->max;
// Again store min by considering
// current node value
var->min = min(var->min, currData);
// Again store max by considering
// current node value
var->max = max(var->max, currData);
return var;
}
void print(Node* root)
{
// Base Case
if (root == NULL)
return;
// Create an empty queue for
// level order traversal
queue<Node*> q;
// Enqueue Root and initialize
// height
q.push(root);
while (q.empty() == false) {
// nodeCount (queue size)
// indicates number
// of nodes at current level.
int nodeCount = q.size();
// Dequeue all nodes of current
// level and Enqueue all nodes
// of next level
while (nodeCount > 0) {
Node* node = q.front();
cout << node->data << " ";
q.pop();
if (node->left != NULL)
q.push(node->left);
if (node->right != NULL)
q.push(node->right);
nodeCount--;
}
}
}
void minMaxProductTree(Node* root)
{
// Utility Function call
minMaxTreeUtil(root);
// Print tree
print(root);
}
// Driver Code
int main()
{
/* 10
/ \
48 3
/ \
11 37
/ \ / \
7 29 42 19
/
7
*/
// Create Binary Tree as shown
Node* root = newNode(10);
root->left = newNode(48);
root->right = newNode(3);
root->right->left = newNode(11);
root->right->right = newNode(37);
root->right->left->left = newNode(7);
root->right->left->right = newNode(29);
root->right->right->left = newNode(42);
root->right->right->right = newNode(19);
root->right->right->right->left = newNode(7);
// Create Min Max Product Tree
minMaxProductTree(root);
return 0;
}
// Java program for the
// above approach
import java.util.*;
// A Tree node
class Node {
public int data;
public Node left, right;
public Node(int x)
{
data = x;
left = null;
right = null;
}
}
// A minMax Structure for storing
// minimum and maximum value
class minMax {
public int min, max;
public minMax(int x, int y)
{
min = x;
max = y;
}
}
class GFG {
// Function to return min value
static int minm(minMax a, minMax b)
{
if (a != null && b != null) {
if (a.min < b.min)
return a.min;
return b.min;
}
if (a == null)
return b.min;
return a.min;
}
// Function to return max value
static int maxm(minMax a, minMax b)
{
if (a != null && b != null) {
if (a.max > b.max)
return a.max;
return b.max;
}
if (a == null)
return b.max;
return a.max;
}
// Utility function to create
// min-max product tree
static minMax minMaxTreeUtil(Node root)
{
// Base condition
if (root == null)
return null;
minMax var1 = new minMax(1000000000, -1000000000);
// Condition to check if
// current node is leaf node
if (root.left == null && root.right == null) {
var1.min = root.data;
var1.max = root.data;
return var1;
}
// Left recursive call
minMax left = minMaxTreeUtil(root.left);
// Right recursive call
minMax right = minMaxTreeUtil(root.right);
// Store Min
var1.min = minm(left, right);
// Store Max
var1.max = maxm(left, right);
// Store current node data
int currData = root.data;
// Assign product of minimum
// and maximum value
root.data = var1.min * var1.max;
// Again store min by considering
// current node value
var1.min = Math.min(var1.min, currData);
// Again store max by considering
// current node value
var1.max = Math.max(var1.max, currData);
return var1;
}
static void printt(Node root)
{
// Base Case
if (root == null)
return;
// Create an empty queue for
// level order traversal
ArrayList<Node> q = new ArrayList<Node>();
// Enqueue Root and initialize
// height
q.add(root);
while (q.size() > 0) {
// nodeCount (queue size)
// indicates number
// of nodes at current level.
int nodeCount = q.size();
// Dequeue all nodes of current
// level and Enqueue all nodes
// of next level
while (nodeCount > 0) {
Node node = q.get(0);
q.remove(0);
System.out.print(node.data + " ");
if (node.left != null)
q.add(node.left);
if (node.right != null)
q.add(node.right);
nodeCount -= 1;
}
}
}
static void minMaxProductTree(Node root)
{
// Utility Function call
minMaxTreeUtil(root);
// Print tree
printt(root);
}
// Driver Code
public static void main(String[] args)
{
// /* 10
// / \
// 48 3
// / \
// 11 37
// / \ / \
// 7 29 42 19
// /
// 7
// */
// Create Binary Tree as shown
Node root = new Node(10);
root.left = new Node(48);
root.right = new Node(3);
root.right.left = new Node(11);
root.right.right = new Node(37);
root.right.left.left = new Node(7);
root.right.left.right = new Node(29);
root.right.right.left = new Node(42);
root.right.right.right = new Node(19);
root.right.right.right.left = new Node(7);
// Create Min Max Product Tree
minMaxProductTree(root);
}
}
// This code is contributed by phasing17
# Python3 program for the
# above approach
from collections import deque
# A Tree node
class Node:
def __init__(self, x):
self.data = x
self.left = None
self.right = None
# A minMax Structure for storing
# minimum and maximum value
class minMax:
def __init__(self, x, y):
self.min = x
self.max = y
# Function to return min value
def minm(a, b):
if (a != None and
b != None):
if a.min < b.min:
return a.min
return b.min
if a == None:
return b.min
return a.min
# Function to return max value
def maxm(a, b):
if a != None and b != None:
if a.max > b.max:
return a.max
return b.max
if a == None:
return b.max
return a.max
# Utility function to create
# min-max product tree
def minMaxTreeUtil(root):
# Base condition
if (root == None):
return None
var = minMax(10 ** 9,
-10 ** 9)
# Condition to check if
# current node is leaf node
if (root.left == None and
root.right == None):
var.min = root.data
var.max = root.data
return var
# Left recursive call
left= minMaxTreeUtil(root.left)
# Right recursive call
right= minMaxTreeUtil(root.right)
# Store Min
var.min = minm(left, right)
# Store Max
var.max = maxm(left, right)
# Store current node data
currData = root.data
# Assign product of minimum
# and maximum value
root.data = var.min * var.max
# Again store min by considering
# current node value
var.min = min(var.min,
currData)
# Again store max by considering
# current node value
var.max = max(var.max,
currData)
return var
def printt(root):
# Base Case
if (root == None):
return
# Create an empty queue for
# level order traversal
q = deque()
# Enqueue Root and initialize
# height
q.append(root)
while (len(q) > 0):
# nodeCount (queue size)
# indicates number
# of nodes at current level.
nodeCount = len(q)
# Dequeue all nodes of current
# level and Enqueue all nodes
# of next level
while (nodeCount > 0):
node = q.popleft()
print(node.data,
end = " ")
if (node.left != None):
q.append(node.left)
if (node.right != None):
q.append(node.right)
nodeCount -= 1
def minMaxProductTree(root):
# Utility Function call
minMaxTreeUtil(root)
# Print tree
printt(root)
# Driver Code
if __name__ == '__main__':
# /* 10
# / \
# 48 3
# / \
# 11 37
# / \ / \
# 7 29 42 19
# /
# 7
# */
# Create Binary Tree as shown
root = Node(10)
root.left = Node(48)
root.right = Node(3)
root.right.left = Node(11)
root.right.right = Node(37)
root.right.left.left = Node(7)
root.right.left.right = Node(29)
root.right.right.left = Node(42)
root.right.right.right = Node(19)
root.right.right.right.left = Node(7)
# Create Min Max Product Tree
minMaxProductTree(root)
# This code is contributed by Mohit Kumar 29
// C# program for the
// above approach
using System;
using System.Collections.Generic;
// A Tree node
class Node {
public int data;
public Node left, right;
public Node(int x)
{
data = x;
left = null;
right = null;
}
}
// A minMax Structure for storing
// minimum and maximum value
class minMax {
public int min, max;
public minMax(int x, int y)
{
min = x;
max = y;
}
}
class GFG {
// Function to return min value
static int minm(minMax a, minMax b)
{
if (a != null && b != null) {
if (a.min < b.min)
return a.min;
return b.min;
}
if (a == null)
return b.min;
return a.min;
}
// Function to return max value
static int maxm(minMax a, minMax b)
{
if (a != null && b != null) {
if (a.max > b.max)
return a.max;
return b.max;
}
if (a == null)
return b.max;
return a.max;
}
// Utility function to create
// min-max product tree
static minMax minMaxTreeUtil(Node root)
{
// Base condition
if (root == null)
return null;
minMax var1 = new minMax(1000000000, -1000000000);
// Condition to check if
// current node is leaf node
if (root.left == null && root.right == null) {
var1.min = root.data;
var1.max = root.data;
return var1;
}
// Left recursive call
minMax left = minMaxTreeUtil(root.left);
// Right recursive call
minMax right = minMaxTreeUtil(root.right);
// Store Min
var1.min = minm(left, right);
// Store Max
var1.max = maxm(left, right);
// Store current node data
int currData = root.data;
// Assign product of minimum
// and maximum value
root.data = var1.min * var1.max;
// Again store min by considering
// current node value
var1.min = Math.Min(var1.min, currData);
// Again store max by considering
// current node value
var1.max = Math.Max(var1.max, currData);
return var1;
}
static void printt(Node root)
{
// Base Case
if (root == null)
return;
// Create an empty queue for
// level order traversal
List<Node> q = new List<Node>();
// Enqueue Root and initialize
// height
q.Add(root);
while (q.Count > 0) {
// nodeCount (queue size)
// indicates number
// of nodes at current level.
int nodeCount = q.Count;
// Dequeue all nodes of current
// level and Enqueue all nodes
// of next level
while (nodeCount > 0) {
Node node = q[0];
q.RemoveAt(0);
Console.Write(node.data + " ");
if (node.left != null)
q.Add(node.left);
if (node.right != null)
q.Add(node.right);
nodeCount -= 1;
}
}
}
static void minMaxProductTree(Node root)
{
// Utility Function call
minMaxTreeUtil(root);
// Print tree
printt(root);
}
// Driver Code
public static void Main(string[] args)
{
// /* 10
// / \
// 48 3
// / \
// 11 37
// / \ / \
// 7 29 42 19
// /
// 7
// */
// Create Binary Tree as shown
Node root = new Node(10);
root.left = new Node(48);
root.right = new Node(3);
root.right.left = new Node(11);
root.right.right = new Node(37);
root.right.left.left = new Node(7);
root.right.left.right = new Node(29);
root.right.right.left = new Node(42);
root.right.right.right = new Node(19);
root.right.right.right.left = new Node(7);
// Create Min Max Product Tree
minMaxProductTree(root);
}
}
// This code is contributed by phasing17
// JS program for the
// above approach
// A Tree node
class Node
{
constructor(x)
{
this.data = x
this.left = null
this.right = null
}
}
// A minMax Structure for storing
// minimum and maximum value
class minMax
{
constructor(x, y)
{
this.min = x
this.max = y
}
}
// Function to return min value
function minm(a, b)
{
if (a != null &&
b != null)
{
if (a.min < b.min)
return a.min
return b.min
}
if (a == null)
return b.min
return a.min
}
// Function to return max value
function maxm(a, b)
{
if (a != null && b != null)
{
if (a.max > b.max)
return a.max
return b.max
}
if (a == null)
return b.max
return a.max
}
// Utility function to create
// min-max product tree
function minMaxTreeUtil(root)
{
// Base condition
if (root == null)
return null
var var1 = new minMax(10 ** 9, - (10 ** 9));
// Condition to check if
// current node is leaf node
if (root.left == null &&
root.right == null)
{
var1.min = root.data
var1.max = root.data
return var1
}
// Left recursive call
left= minMaxTreeUtil(root.left)
// Right recursive call
right= minMaxTreeUtil(root.right)
// Store Min
var1.min = minm(left, right)
// Store Max
var1.max = maxm(left, right)
// Store current node data
currData = root.data
// Assign product of minimum
// and maximum value
root.data = var1.min * var1.max
// Again store min by considering
// current node value
var1.min = Math.min(var1.min,
currData)
// Again store max by considering
// current node value
var1.max = Math.max(var1.max,
currData)
return var1
}
function printt(root)
{
// Base Case
if (root == null)
return
// Create an empty queue for
// level order traversal
q = []
// Enqueue Root and initialize
// height
q.push(root)
while (q.length > 0)
{
// nodeCount (queue size)
// indicates number
// of nodes at current level.
nodeCount = q.length
// Dequeue all nodes of current
// level and Enqueue all nodes
// of next level
while (nodeCount > 0)
{
node = q.shift()
process.stdout.write(node.data + " ")
if (node.left != null)
q.push(node.left)
if (node.right != null)
q.push(node.right)
nodeCount -= 1
}
}
}
function minMaxProductTree(root)
{
// Utility Function call
minMaxTreeUtil(root)
// Print tree
printt(root)
}
// Driver Code
// /* 10
// / \
// 48 3
// / \
// 11 37
// / \ / \
// 7 29 42 19
// /
// 7
// */
// Create Binary Tree as shown
root = new Node(10)
root.left = new Node(48)
root.right = new Node(3)
root.right.left = new Node(11)
root.right.right = new Node(37)
root.right.left.left = new Node(7)
root.right.left.right = new Node(29)
root.right.right.left = new Node(42)
root.right.right.right =new Node(19)
root.right.right.right.left = new Node(7)
// Create Min Max Product Tree
minMaxProductTree(root)
// This code is contributed by phasing17
Output: 144 48 294 203 294 7 29 42 49 7
Time Complexity: O(N), where N denotes the number of nodes in the binary tree
Auxiliary Space: O(N)
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