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Find maximum (or minimum) in Binary Tree

Get maximum left node in binary tree

Last Updated : 28 Mar, 2023

Given a tree, the task is to find the maximum in an only left node of the binary tree.

Examples:

Input :             7          /    \         6       5        / \     / \       4  3     2  1         Output : 6  Input :             1          /    \         2       3        /       / \       4       5   6         \    /  \           7  8   9            Output : 8

Traverse with inorder traversal and Apply the condition for the left node only and get maximum of left node.

Implementation: Let’s try to understand with code.

C++

   // CPP program to print maximum element
// in left node.
#include <bits/stdc++.h>
using namespace std;
 
// A Binary Tree Node
struct Node {
    int data;
    struct Node *left, *right;
};
 
// Get max of left element using
// Inorder traversal
int maxOfLeftElement(Node* root)
{
    int res = INT_MIN;
    if (root == NULL)
        return res;
 
    if (root->left != NULL)
        res = root->left->data;
 
    // Return maximum of three values
    // 1) Recursive max in left subtree
    // 2) Value in left node
    // 3) Recursive max in right subtree
    return max({ maxOfLeftElement(root->left),
                 res,
                 maxOfLeftElement(root->right) });
}
 
// Utility function to create a new tree node
Node* newNode(int data)
{
    Node* temp = new Node;
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}
 
// Driver program to test above functions
int main()
{
    // Let us create binary tree shown in above diagram
    Node* root = newNode(7);
    root->left = newNode(6);
    root->right = newNode(5);
    root->left->left = newNode(4);
    root->left->right = newNode(3);
    root->right->left = newNode(2);
    root->right->right = newNode(1);
 
    /*     7
         /    \
        6       5
       / \     / \
      4  3     2  1          */
    cout << maxOfLeftElement(root);
    return 0;
}
 
  

Java

   // Java program to print maximum element 
// in left node. 
import java.util.*;
class GfG {
 
// A Binary Tree Node 
static class Node { 
    int data; 
    Node left, right; 
}
 
// Get max of left element using 
// Inorder traversal 
static int maxOfLeftElement(Node root) 
{ 
    int res = Integer.MIN_VALUE; 
    if (root == null) 
        return res; 
 
    if (root.left != null) 
        res = root.left.data; 
 
    // Return maximum of three values 
    // 1) Recursive max in left subtree 
    // 2) Value in left node 
    // 3) Recursive max in right subtree 
    return Math.max(maxOfLeftElement(root.left),
       Math.max(res, maxOfLeftElement(root.right))); 
} 
 
// Utility function to create a new tree node 
static Node newNode(int data) 
{ 
    Node temp = new Node(); 
    temp.data = data; 
    temp.left = null;
    temp.right = null; 
    return temp; 
} 
 
// Driver program to test above functions 
public static void main(String[] args) 
{ 
    // Let us create binary tree shown in above diagram 
    Node root = newNode(7); 
    root.left = newNode(6); 
    root.right = newNode(5); 
    root.left.left = newNode(4); 
    root.left.right = newNode(3); 
    root.right.left = newNode(2); 
    root.right.right = newNode(1); 
 
    /* 7 
        / \ 
        6 5 
    / \ / \ 
    4 3 2 1         */
    System.out.println(maxOfLeftElement(root)); 
}
} 
 
  

Python3

   # Python program to print maximum element 
# in left node. 
 
# Utility class to create a 
# new tree node 
class newNode:
    def __init__(self, data): 
        self.data = data 
        self.left = self.right = None
     
# Get max of left element using 
# Inorder traversal 
def maxOfLeftElement(root):
    res = -999999999999
    if (root == None):
        return res 
 
    if (root.left != None): 
        res = root.left.data 
 
    # Return maximum of three values 
    # 1) Recursive max in left subtree 
    # 2) Value in left node 
    # 3) Recursive max in right subtree 
    return max({ maxOfLeftElement(root.left), res,
                 maxOfLeftElement(root.right) })
 
# Driver Code
if __name__ == '__main__':
 
    # Let us create binary tree shown
    # in above diagram 
    root = newNode(7) 
    root.left = newNode(6) 
    root.right = newNode(5) 
    root.left.left = newNode(4) 
    root.left.right = newNode(3) 
    root.right.left = newNode(2) 
    root.right.right = newNode(1) 
 
    #     7 
    #     / \ 
    # 6     5 
    # / \     / \ 
    # 4 3     2 1         
    print(maxOfLeftElement(root))
 
# This code is contributed by PranchalK
 
  

C#

   // C# program to print maximum element 
// in left node.
using System;
 
class GfG 
{
 
    // A Binary Tree Node 
    class Node 
    { 
        public int data; 
        public Node left, right; 
    }
 
    // Get max of left element using 
    // Inorder traversal 
    static int maxOfLeftElement(Node root) 
    { 
        int res = int.MinValue; 
        if (root == null) 
            return res; 
 
        if (root.left != null) 
            res = root.left.data; 
 
        // Return maximum of three values 
        // 1) Recursive max in left subtree 
        // 2) Value in left node 
        // 3) Recursive max in right subtree 
        return Math.Max(maxOfLeftElement(root.left),
        Math.Max(res, maxOfLeftElement(root.right))); 
    } 
 
    // Utility function to create a new tree node 
    static Node newNode(int data) 
    { 
        Node temp = new Node(); 
        temp.data = data; 
        temp.left = null;
        temp.right = null; 
        return temp; 
    } 
 
    // Driver code 
    public static void Main(String[] args) 
    { 
        // Let us create binary tree
        // shown in above diagram 
        Node root = newNode(7); 
        root.left = newNode(6); 
        root.right = newNode(5); 
        root.left.left = newNode(4); 
        root.left.right = newNode(3); 
        root.right.left = newNode(2); 
        root.right.right = newNode(1); 
 
        /* 7 
            / \ 
            6 5 
        / \ / \ 
        4 3 2 1         */
        Console.WriteLine(maxOfLeftElement(root)); 
    }
}
 
// This code is contributed by 29AjayKumar
 
  

Javascript

   <script>
 
// JavaScript program to print maximum element 
// in left node.
// A Binary Tree Node 
class Node 
{ 
  constructor()
  {
    this.data = 0;
    this.left = null;
    this.right = null;
  }
}
// Get max of left element using 
// Inorder traversal 
function maxOfLeftElement(root) 
{ 
    var res = -1000000000; 
    if (root == null) 
        return res; 
    if (root.left != null) 
        res = root.left.data; 
    // Return maximum of three values 
    // 1) Recursive max in left subtree 
    // 2) Value in left node 
    // 3) Recursive max in right subtree 
    return Math.max(maxOfLeftElement(root.left),
    Math.max(res, maxOfLeftElement(root.right))); 
} 
// Utility function to create a new tree node 
function newNode(data) 
{ 
    var temp = new Node(); 
    temp.data = data; 
    temp.left = null;
    temp.right = null; 
    return temp; 
} 
// Driver code 
// Let us create binary tree
// shown in above diagram 
var root = newNode(7); 
root.left = newNode(6); 
root.right = newNode(5); 
root.left.left = newNode(4); 
root.left.right = newNode(3); 
root.right.left = newNode(2); 
root.right.right = newNode(1); 
/* 7 
    / \ 
    6 5 
/ \ / \ 
4 3 2 1         */
document.write(maxOfLeftElement(root)); 
 
</script>
 
  

Output

Iterative Approach(using queue):
Follow the below steps to solve the given problem:
1). Perform level order traversal using queue data structure.
2). At each node check it’s left children is null or not. If the left children is not null then compare its with the existing max left value.
3). If the current node left child value is greater than the existing value then update the max left value with current node left child value.

Below is the implementation of above approach:

C++

   // C++ Program to print maximum element
// in left node
#include<bits/stdc++.h>
using namespace std;
 
// a binary tree node
struct Node{
    int data;
    Node* left;
    Node* right;
};
 
// utiltity function to create a new tree node
Node* newNode(int data){
    Node* temp = new Node;
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}
 
// get max of left element by
// level order traversal using queue
int maxOfLeftElement(Node* root){
    int res = INT_MIN;
    if(root == NULL) return res;
    queue<Node*> q;
    q.push(root);
    while(!q.empty()){
        Node* front_node = q.front();
        q.pop();
        if(front_node->left != NULL){
            res = max(res, front_node->left->data);
        }
        if(front_node->left) q.push(front_node->left);
        if(front_node->right) q.push(front_node->right);
    }
    return res;
}
 
// Driver program to test above functions
int main(){
    // Let us create binary tree shown in above diagram
    Node* root = newNode(7);
    root->left = newNode(6);
    root->right = newNode(5);
    root->left->left = newNode(4);
    root->left->right = newNode(3);
    root->right->left = newNode(2);
    root->right->right = newNode(1);
  
    /*     7
         /    \
        6       5
       / \     / \
      4  3     2  1          */
    cout << maxOfLeftElement(root);
    return 0;
}
 
// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002)
 
  

Java

   // Java program for the above approach
import java.util.*;
 
// a binary tree node
class Node {
    int data;
    Node left;
    Node right;
 
    Node(int data) {
        this.data = data;
        left = right = null;
    }
}
 
class Main {
 
    // get max of left element by
    // level order traversal using queue
    static int maxOfLeftElement(Node root) {
        int res = Integer.MIN_VALUE;
        if(root == null) return res;
        Queue<Node> q = new LinkedList<Node>();
        q.offer(root);
        while(!q.isEmpty()){
            Node front_node = q.poll();
            if(front_node.left != null){
                res = Math.max(res, front_node.left.data);
            }
            if(front_node.left != null) q.offer(front_node.left);
            if(front_node.right != null) q.offer(front_node.right);
        }
        return res;
    }
 
    // Driver program to test above functions
    public static void main(String[] args) {
        // Let us create binary tree shown in above diagram
        Node root = new Node(7);
        root.left = new Node(6);
        root.right = new Node(5);
        root.left.left = new Node(4);
        root.left.right = new Node(3);
        root.right.left = new Node(2);
        root.right.right = new Node(1);
 
        /*     7
             /    \
            6       5
           / \     / \
          4  3     2  1          */
        System.out.println(maxOfLeftElement(root));
    }
}
 
// This code is contributed by codebraxnzt
 
  

Python3

   from queue import Queue
 
# a binary tree node
class Node:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
 
# get max of left element by level order traversal using queue
def maxOfLeftElement(root):
    res = float('-inf')
    if root is None:
        return res
    q = Queue()
    q.put(root)
    while not q.empty():
        front_node = q.get()
        if front_node.left:
            res = max(res, front_node.left.data)
        if front_node.left:
            q.put(front_node.left)
        if front_node.right:
            q.put(front_node.right)
    return res
 
# Driver program to test above functions
if __name__ == '__main__':
    # Let us create binary tree shown in above diagram
    root = Node(7)
    root.left = Node(6)
    root.right = Node(5)
    root.left.left = Node(4)
    root.left.right = Node(3)
    root.right.left = Node(2)
    root.right.right = Node(1)
 
    """
         7
        / \
       6   5
      / \ / \
     4  3 2  1
    """
    print(maxOfLeftElement(root))
 
  

C#

   // C# Program to print maximum element
// in left node
using System;
using System.Collections.Generic;
 
// a binary tree node
class Node{
    public int data;
    public Node left;
    public Node right;
 
    public Node(int data){
        this.data = data;
        left = null;
        right = null;
    }
}
 
class BinaryTree
{
   
    // get max of left element by
    // level order traversal using queue
    public static int maxOfLeftElement(Node root){
        int res = int.MinValue;
        if(root == null) return res;
        Queue<Node> q = new Queue<Node>();
        q.Enqueue(root);
        while(q.Count > 0){
            Node front_node = q.Dequeue();
            if(front_node.left != null){
                res = Math.Max(res, front_node.left.data);
            }
            if(front_node.left != null) q.Enqueue(front_node.left);
            if(front_node.right != null) q.Enqueue(front_node.right);
        }
        return res;
    }
 
    // Driver program to test above functions
    static void Main(){
        // Let us create binary tree shown in above diagram
        Node root = new Node(7);
        root.left = new Node(6);
        root.right = new Node(5);
        root.left.left = new Node(4);
        root.left.right = new Node(3);
        root.right.left = new Node(2);
        root.right.right = new Node(1);
  
        /*     7
             /    \
            6       5
           / \     / \
          4  3     2  1          */
        Console.WriteLine(maxOfLeftElement(root));
    }
}
 
  

Javascript

   // javascript Program to print maximum element
// in left node
 
// a binary tree node
class Node{
     
    constructor(){
        this.data = 0;
        this.left = null;
        this.right = null;
    }
}
 
// utiltity function to create a new tree node
function newNode(data){
    let temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
}
 
// get max of left element by
// level order traversal using queue
function maxOfLeftElement(root){
    let res = -2000;
    if(root == null) return res;
    let q = [];
    q.push(root);
     
    while(q.length > 0){
        let front_node = q[0];
        q.shift();
        if(front_node.left != null){
            res = Math.max(res, front_node.left.data);
        }
        if(front_node.left) q.push(front_node.left);
        if(front_node.right) q.push(front_node.right);
    }
    return res;
}
 
// Driver program to test above functions
// Let us create binary tree shown in above diagram
let root = newNode(7);
root.left = newNode(6);
root.right = newNode(5);
root.left.left = newNode(4);
root.left.right = newNode(3);
root.right.left = newNode(2);
root.right.right = newNode(1);
 
/*     7
     /    \
    6       5
   / \     / \
  4  3     2  1          */
console.log(maxOfLeftElement(root));
 
// The code is contributed by Nidhi goel.
 
  

Output

Time Complexity: O(N) where N is the number of nodes in given binary tree.
Auxiliary Space: O(N) due to queue data structure.

Find maximum (or minimum) in Binary Tree

devanshuagarwal

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Article Tags :

DSA
Tree

Practice Tags :

Tree

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