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Find maximum and minimum element in binary tree without using recursion or stack or queue

  • Last Updated : 04 Aug, 2021

Given a binary tree. The task is to find out the maximum and minimum element in a binary tree without using recursion or stack or queue i.e, space complexity should be O(1). 

Examples: 

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Input : 
                       12
                     /     \
                   13       10
                          /     \
                       14       15
                      /   \     /  \
                     21   24   22   23

Output : Max element : 24
         Min element : 10

Input : 
                       12
                     /     \
                  19        82
                 /        /     \
               41       15       95
                 \     /         /  \
                  2   21        7   16

Output : Max element : 95
         Min element : 2

Prerequisite : Inorder Tree Traversal without recursion and without stack



Approach : 
1. Initialize current as root 
2. Take to variable max and min 
3. While current is not NULL 

  • If the current does not have left child 
    • Update variable max and min with current’s data if required
    • Go to the right, i.e., current = current->right
  • Else 
    • Make current as the right child of the rightmost 
      node in current’s left subtree
    • Go to this left child, i.e., current = current->left

Below is the implementation of the above approach :  

C++




// C++ program find maximum and minimum element
#include <bits/stdc++.h>
using namespace std;
 
// A Tree node
struct Node {
    int key;
    struct Node *left, *right;
};
 
// Utility function to create a new node
Node* newNode(int key)
{
    Node* temp = new Node;
    temp->key = key;
    temp->left = temp->right = NULL;
    return (temp);
}
 
 
// Function to print a maximum and minimum element
// in a tree without recursion without stack
void printMinMax(Node* root)
{
     
    if (root == NULL)
    {
        cout << "Tree is empty";
        return;
    }
     
    Node* current = root;
     
    Node* pre;
     
    // Max variable for storing maximum value   
    int max_value = INT_MIN;
     
    // Min variable for storing minimum value   
    int min_value = INT_MAX;
     
     
    while (current != NULL)
    {
        // If left child does nor exists
        if (current->left == NULL)
        {
            max_value = max(max_value, current->key);
            min_value = min(min_value, current->key);
             
            current = current->right;
        }
        else
        {
   
            // Find the inorder predecessor of current
            pre = current->left;
            while (pre->right != NULL && pre->right !=
                                                 current)
                pre = pre->right;
   
            // Make current as the right child
            // of its inorder predecessor
            if (pre->right == NULL)
            {
                pre->right = current;
                current = current->left;
            }
   
            // Revert the changes made in the 'if' part to
            // restore the original tree i.e., fix the
            // right child of predecessor
            else
            {
                pre->right = NULL;
                 
                max_value = max(max_value, current->key);
                min_value = min(min_value, current->key);
             
                current = current->right;
            } // End of if condition pre->right == NULL
             
        } // End of if condition current->left == NULL
         
    } // End of while
     
    // Finally print max and min value
    cout << "Max Value is : " << max_value << endl;
    cout << "Min Value is : " << min_value << endl;
}
 
// Driver Code
int main()
{
    /* 15
      /  \
    19   11
        /  \
       25   5
      / \   / \
    17  3  23  24
 
    Let us create Binary Tree as shown
    above */
 
    Node* root = newNode(15);
    root->left = newNode(19);
    root->right = newNode(11);
 
    root->right->left = newNode(25);
    root->right->right = newNode(5);
 
    root->right->left->left = newNode(17);
    root->right->left->right = newNode(3);
    root->right->right->left = newNode(23);
    root->right->right->right = newNode(24);
     
    // Function call for printing a max
    // and min element in a tree
    printMinMax(root);
 
    return 0;
}

Java




// Java program find maximum and minimum element
class GFG
{
 
// A Tree node
static class Node
{
    int key;
    Node left, right;
};
 
// Utility function to create a new node
static Node newNode(int key)
{
    Node temp = new Node();
    temp.key = key;
    temp.left = temp.right = null;
    return (temp);
}
 
 
// Function to print a maximum and minimum element
// in a tree without recursion without stack
static void printMinMax(Node root)
{
     
    if (root == null)
    {
        System.out.print("Tree is empty");
        return;
    }
     
    Node current = root;
     
    Node pre;
     
    // Max variable for storing maximum value
    int max_value = Integer.MIN_VALUE;
     
    // Min variable for storing minimum value
    int min_value = Integer.MAX_VALUE;
     
     
    while (current != null)
    {
        // If left child does nor exists
        if (current.left == null)
        {
            max_value = Math.max(max_value, current.key);
            min_value = Math.min(min_value, current.key);
             
            current = current.right;
        }
        else
        {
     
            // Find the inorder predecessor of current
            pre = current.left;
            while (pre.right != null && pre.right !=
                                                current)
                pre = pre.right;
     
            // Make current as the right child
            // of its inorder predecessor
            if (pre.right == null)
            {
                pre.right = current;
                current = current.left;
            }
     
            // Revert the changes made in the 'if' part to
            // restore the original tree i.e., fix the
            // right child of predecessor
            else
            {
                pre.right = null;
                 
                max_value = Math.max(max_value, current.key);
                min_value = Math.min(min_value, current.key);
             
                current = current.right;
            } // End of if condition pre.right == null
             
        } // End of if condition current.left == null
         
    } // End of while
     
    // Finally print max and min value
    System.out.print("Max Value is : " + max_value + "\n");
    System.out.print("Min Value is : " + min_value + "\n");
}
 
// Driver Code
public static void main(String[] args)
{
    /* 15
    / \
    19 11
        / \
    25 5
    / \ / \
    17 3 23 24
 
    Let us create Binary Tree as shown
    above */
 
    Node root = newNode(15);
    root.left = newNode(19);
    root.right = newNode(11);
 
    root.right.left = newNode(25);
    root.right.right = newNode(5);
 
    root.right.left.left = newNode(17);
    root.right.left.right = newNode(3);
    root.right.right.left = newNode(23);
    root.right.right.right = newNode(24);
     
    // Function call for printing a max
    // and min element in a tree
    printMinMax(root);
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python program find maximum and minimum element
from sys import maxsize
 
INT_MAX = maxsize
INT_MIN = -maxsize
 
# A Tree node
class Node:
    def __init__(self, key):
        self.key = key
        self.left = None
        self.right = None
 
# Function to print a maximum and minimum element
# in a tree without recursion without stack
def printMinMax(root: Node):
    if root is None:
        print("Tree is empty")
        return
 
    current = root
    pre = Node(0)
 
    # Max variable for storing maximum value
    max_value = INT_MIN
 
    # Min variable for storing minimum value
    min_value = INT_MAX
 
    while current is not None:
 
        # If left child does nor exists
        if current.left is None:
            max_value = max(max_value, current.key)
            min_value = min(min_value, current.key)
 
            current = current.right
        else:
 
            # Find the inorder predecessor of current
            pre = current.left
            while pre.right is not None and pre.right != current:
                pre = pre.right
 
            # Make current as the right child
            # of its inorder predecessor
            if pre.right is None:
                pre.right = current
                current = current.left
 
            # Revert the changes made in the 'if' part to
            # restore the original tree i.e., fix the
            # right child of predecessor
            else:
                pre.right = None
                max_value = max(max_value, current.key)
                min_value = min(min_value, current.key)
 
                current = current.right
 
            # End of if condition pre->right == NULL
 
        # End of if condition current->left == NULL
 
    # End of while
 
    # Finally print max and min value
    print("Max value is :", max_value)
    print("Min value is :", min_value)
 
# Driver Code
if __name__ == "__main__":
 
    # /* 15
    # / \
    # 19 11
    #     / \
    # 25 5
    # / \ / \
    # 17 3 23 24
 
    # Let us create Binary Tree as shown
    # above */
 
    root = Node(15)
    root.left = Node(19)
    root.right = Node(11)
 
    root.right.left = Node(25)
    root.right.right = Node(5)
 
    root.right.left.left = Node(17)
    root.right.left.right = Node(3)
    root.right.right.left = Node(23)
    root.right.right.right = Node(24)
 
    # Function call for printing a max
    # and min element in a tree
    printMinMax(root)
 
# This code is contributed by
# sanjeev2552

C#




// C# program find maximum and minimum element
using System;
class GFG
{
 
// A Tree node
class Node
{
    public int key;
    public Node left, right;
};
 
// Utility function to create a new node
static Node newNode(int key)
{
    Node temp = new Node();
    temp.key = key;
    temp.left = temp.right = null;
    return (temp);
}
 
// Function to print a maximum and minimum element
// in a tree without recursion without stack
static void printMinMax(Node root)
{
    if (root == null)
    {
        Console.Write("Tree is empty");
        return;
    }
     
    Node current = root;
     
    Node pre;
     
    // Max variable for storing maximum value
    int max_value = int.MinValue;
     
    // Min variable for storing minimum value
    int min_value = int.MaxValue;
     
    while (current != null)
    {
        // If left child does nor exists
        if (current.left == null)
        {
            max_value = Math.Max(max_value,
                                 current.key);
            min_value = Math.Min(min_value,
                                 current.key);
             
            current = current.right;
        }
        else
        {
     
            // Find the inorder predecessor of current
            pre = current.left;
            while (pre.right != null &&
                   pre.right != current)
                pre = pre.right;
     
            // Make current as the right child
            // of its inorder predecessor
            if (pre.right == null)
            {
                pre.right = current;
                current = current.left;
            }
     
            // Revert the changes made in the 'if' part to
            // restore the original tree i.e., fix the
            // right child of predecessor
            else
            {
                pre.right = null;
                 
                max_value = Math.Max(max_value,
                                     current.key);
                min_value = Math.Min(min_value,    
                                     current.key);
             
                current = current.right;
            } // End of if condition pre.right == null
             
        } // End of if condition current.left == null
         
    } // End of while
     
    // Finally print max and min value
    Console.Write("Max Value is : " +
                   max_value + "\n");
    Console.Write("Min Value is : " +
                   min_value + "\n");
}
 
// Driver Code
public static void Main(String[] args)
{
    /* 15
    / \
    19 11
        / \
    25 5
    / \ / \
    17 3 23 24
 
    Let us create Binary Tree as shown
    above */
 
    Node root = newNode(15);
    root.left = newNode(19);
    root.right = newNode(11);
 
    root.right.left = newNode(25);
    root.right.right = newNode(5);
 
    root.right.left.left = newNode(17);
    root.right.left.right = newNode(3);
    root.right.right.left = newNode(23);
    root.right.right.right = newNode(24);
     
    // Function call for printing a max
    // and min element in a tree
    printMinMax(root);
}
}
 
// This code is contributed by PrinciRaj1992

Javascript




<script>
 
// Javascript program find maximum
// and minimum element
 
// A Tree node
class Node
{
    constructor()
    {
        this.key = 0;
        this.left = null;
        this.right = null;
    }
};
 
// Utility function to create a new node
function newNode(key)
{
    var temp = new Node();
    temp.key = key;
    temp.left = temp.right = null;
    return (temp);
}
 
// Function to print a maximum and minimum
// element in a tree without recursion
// without stack
function printMinMax(root)
{
    if (root == null)
    {
        document.write("Tree is empty");
        return;
    }
     
    var current = root;
    var pre;
     
    // Max variable for storing maximum value
    var max_value = -1000000000;
     
    // Min variable for storing minimum value
    var min_value = 1000000000;
     
    while (current != null)
    {
         
        // If left child does nor exists
        if (current.left == null)
        {
            max_value = Math.max(max_value,
                                 current.key);
            min_value = Math.min(min_value,
                                 current.key);
             
            current = current.right;
        }
        else
        {
             
            // Find the inorder predecessor of current
            pre = current.left;
            while (pre.right != null &&
                   pre.right != current)
                pre = pre.right;
     
            // Make current as the right child
            // of its inorder predecessor
            if (pre.right == null)
            {
                pre.right = current;
                current = current.left;
            }
     
            // Revert the changes made in the 'if'
            // part to restore the original tree
            // i.e., fix the right child of predecessor
            else
            {
                pre.right = null;
                 
                max_value = Math.max(max_value,
                                     current.key);
                min_value = Math.min(min_value,    
                                     current.key);
             
                current = current.right;
            } // End of if condition pre.right == null
             
        } // End of if condition current.left == null
         
    } // End of while
     
    // Finally print max and min value
    document.write("Max Value is : " +
                   max_value + "<br>");
    document.write("Min Value is : " +
                   min_value + "<br>");
}
 
// Driver Code
/* 15
/ \
19 11
    / \
25 5
/ \ / \
17 3 23 24
Let us create Binary Tree as shown
above */
var root = newNode(15);
root.left = newNode(19);
root.right = newNode(11);
root.right.left = newNode(25);
root.right.right = newNode(5);
root.right.left.left = newNode(17);
root.right.left.right = newNode(3);
root.right.right.left = newNode(23);
root.right.right.right = newNode(24);
 
// Function call for printing a max
// and min element in a tree
printMinMax(root);
 
// This code is contributed by noob2000
 
</script>

Output : 

Max Value is : 25
Min Value is : 3

Time Complexity: O(N)
Space complexity: O(1)
 




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