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

Last Updated : 04 Aug, 2021
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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: 

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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