Find maximum among all left nodes in Binary Tree

Given a binary tree that contains nodes of integer values. Our goal is to find the largest left node among all the node which is exist in given tree. For example.

Largest left node in binary tree

In above tree the left largest node is 11. In case tree is empty or tree have the no right child then result is None. Here given code implementation process.

// Java program for
// Find maximum among all left nodes in Binary Tree 
class TreeNode
{
    // Data value 
    public int data;
    // Indicates left and right subtree
    public TreeNode left;
    public TreeNode right;
    public TreeNode(int data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
public class BinaryTree
{
    public TreeNode root;
    public TreeNode result;
    public BinaryTree()
    {
        this.root = null;
        this.result = null;
    }
    public void findMaxLeftNode(TreeNode node)
    {
        if (node != null)
        {
            if (node.left != null)
            {
                if (this.result == null || 
                    this.result.data < node.left.data)
                {
                    // Get new max left child node
                    this.result = node.left;
                }
            }
            // Visit left subtree
            findMaxLeftNode(node.left);
            // Visit right subtree
            findMaxLeftNode(node.right);
        }
    }
    public void maxLeftChild()
    {
        this.result = null;
        this.findMaxLeftNode(this.root);
        if (this.result == null)
        {
            // When no left child node
            System.out.println("None");
        }
        else
        {
            // Display calculated result
            System.out.println(this.result.data);
        }
    }
    public static void main(String[] args)
    {
        BinaryTree tree = new BinaryTree();
        /*  
            Binary Tree
            ---------------
                  1
                 /  \
                11   3
               /    /  \
              4    5    6   
             / \    \    \  
            8   7    12   -4
               / \       
             -2   9    
        */
        // Add Binary tree nodes
        tree.root = new TreeNode(1);
        tree.root.left = new TreeNode(11);
        tree.root.right = new TreeNode(3);
        tree.root.right.right = new TreeNode(6);
        tree.root.right.right.right = new TreeNode(-4);
        tree.root.right.left = new TreeNode(5);
        tree.root.left.left = new TreeNode(4);
        tree.root.left.left.left = new TreeNode(8);
        tree.root.left.left.right = new TreeNode(7);
        tree.root.left.left.right.left = new TreeNode(-2);
        tree.root.left.left.right.right = new TreeNode(9);
        tree.root.right.left.right = new TreeNode(12);
        /*
                  1          
                 /  \
              → 11   3       
               /    /  \
              4    5    6    
             / \    \    \   
            8   7    12   -4  
               / \       
             -2   9      

            --------------------
            Max left node is : 11
        */
        tree.maxLeftChild();
    }
}

Output

11
// Include header file
#include <iostream>

using namespace std;
// C++ program for
// Find maximum among all left nodes in Binary Tree
class TreeNode
{
    public:
    // Data value
    int data;
    // Indicates left and right subtree
    TreeNode *left;
    TreeNode *right;
    TreeNode(int data)
    {
        this->data = data;
        this->left = NULL;
        this->right = NULL;
    }
};
class BinaryTree
{
    public: TreeNode *root;
    TreeNode *result;
    BinaryTree()
    {
        this->root = NULL;
        this->result = NULL;
    }
    void findMaxLeftNode(TreeNode *node)
    {
        if (node != NULL)
        {
            if (node->left != NULL)
            {
                if (this->result == NULL || 
                    this->result->data < node->left->data)
                {
                    // Get new max left child node
                    this->result = node->left;
                }
            }
            // Visit left subtree
            this->findMaxLeftNode(node->left);
            // Visit right subtree
            this->findMaxLeftNode(node->right);
        }
    }
    void maxLeftChild()
    {
        this->result = NULL;
        this->findMaxLeftNode(this->root);
        if (this->result == NULL)
        {
            // When no left child node
            cout << "None" << endl;
        }
        else
        {
            // Display calculated result
            cout << this->result->data << endl;
        }
    }
};
int main()
{
    BinaryTree *tree = new BinaryTree();
    /*
        Binary Tree
        ---------------
              1
             /  \
            11   3
           /    /  \
          4    5    6   
         / \    \    \  
        8   7    12   -4
           / \       
         -2   9    
    */
    // Add Binary tree nodes
    tree->root = new TreeNode(1);
    tree->root->left = new TreeNode(11);
    tree->root->right = new TreeNode(3);
    tree->root->right->right = new TreeNode(6);
    tree->root->right->right->right = new TreeNode(-4);
    tree->root->right->left = new TreeNode(5);
    tree->root->left->left = new TreeNode(4);
    tree->root->left->left->left = new TreeNode(8);
    tree->root->left->left->right = new TreeNode(7);
    tree->root->left->left->right->left = new TreeNode(-2);
    tree->root->left->left->right->right = new TreeNode(9);
    tree->root->right->left->right = new TreeNode(12);
    /*
              1          
             /  \
          → 11   3       
           /    /  \
          4    5    6    
         / \    \    \   
        8   7    12   -4  
           / \       
         -2   9      
        --------------------
        Max left node is : 11
    */
    tree->maxLeftChild();
    return 0;
}

Output

11
package main
import "fmt"
// Go program for
// Find maximum among all left nodes in Binary Tree
type TreeNode struct {
    // Data value
    data int
    // Indicates left and right subtree
    left * TreeNode
    right * TreeNode
}
func getTreeNode(data int) * TreeNode {
    var me *TreeNode = &TreeNode {}
    me.data = data
    me.left = nil
    me.right = nil
    return me
}
type BinaryTree struct {
    root * TreeNode
    result * TreeNode
}
func getBinaryTree() * BinaryTree {
    var me *BinaryTree = &BinaryTree {}
    me.root = nil
    me.result = nil
    return me
}
func(this *BinaryTree) findMaxLeftNode(node * TreeNode) {
    if node != nil {
        if node.left != nil {
            if this.result == nil || this.result.data < node.left.data {
                // Get new max left child node
                this.result = node.left
            }
        }
        // Visit left subtree
        this.findMaxLeftNode(node.left)
        // Visit right subtree
        this.findMaxLeftNode(node.right)
    }
}
func(this BinaryTree) maxLeftChild() {
    this.result = nil
    this.findMaxLeftNode(this.root)
    if this.result == nil {
        // When no left child node
        fmt.Println("None")
    } else {
        // Display calculated result
        fmt.Println(this.result.data)
    }
}
func main() {
    var tree * BinaryTree = getBinaryTree()
    /*
        Binary Tree
        ---------------
              1
             /  \
            11   3
           /    /  \
          4    5    6   
         / \    \    \  
        8   7    12   -4
           / \       
         -2   9    
    */
    // Add Binary tree nodes
    tree.root = getTreeNode(1)
    tree.root.left = getTreeNode(11)
    tree.root.right = getTreeNode(3)
    tree.root.right.right = getTreeNode(6)
    tree.root.right.right.right = getTreeNode(-4)
    tree.root.right.left = getTreeNode(5)
    tree.root.left.left = getTreeNode(4)
    tree.root.left.left.left = getTreeNode(8)
    tree.root.left.left.right = getTreeNode(7)
    tree.root.left.left.right.left = getTreeNode(-2)
    tree.root.left.left.right.right = getTreeNode(9)
    tree.root.right.left.right = getTreeNode(12)
    /*
              1          
             /  \
          → 11   3       
           /    /  \
          4    5    6    
         / \    \    \   
        8   7    12   -4  
           / \       
         -2   9      
        --------------------
        Max left node is : 11
    */
    tree.maxLeftChild()
}

Output

11
// Include namespace system
using System;
// Csharp program for
// Find maximum among all left nodes in Binary Tree
public class TreeNode
{
    // Data value
    public int data;
    // Indicates left and right subtree
    public TreeNode left;
    public TreeNode right;
    public TreeNode(int data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
public class BinaryTree
{
    public TreeNode root;
    public TreeNode result;
    public BinaryTree()
    {
        this.root = null;
        this.result = null;
    }
    public void findMaxLeftNode(TreeNode node)
    {
        if (node != null)
        {
            if (node.left != null)
            {
                if (this.result == null || 
                    this.result.data < node.left.data)
                {
                    // Get new max left child node
                    this.result = node.left;
                }
            }
            // Visit left subtree
            this.findMaxLeftNode(node.left);
            // Visit right subtree
            this.findMaxLeftNode(node.right);
        }
    }
    public void maxLeftChild()
    {
        this.result = null;
        this.findMaxLeftNode(this.root);
        if (this.result == null)
        {
            // When no left child node
            Console.WriteLine("None");
        }
        else
        {
            // Display calculated result
            Console.WriteLine(this.result.data);
        }
    }
    public static void Main(String[] args)
    {
        BinaryTree tree = new BinaryTree();
        /*
            Binary Tree
            ---------------
                  1
                 /  \
                11   3
               /    /  \
              4    5    6   
             / \    \    \  
            8   7    12   -4
               / \       
             -2   9    
        */
        // Add Binary tree nodes
        tree.root = new TreeNode(1);
        tree.root.left = new TreeNode(11);
        tree.root.right = new TreeNode(3);
        tree.root.right.right = new TreeNode(6);
        tree.root.right.right.right = new TreeNode(-4);
        tree.root.right.left = new TreeNode(5);
        tree.root.left.left = new TreeNode(4);
        tree.root.left.left.left = new TreeNode(8);
        tree.root.left.left.right = new TreeNode(7);
        tree.root.left.left.right.left = new TreeNode(-2);
        tree.root.left.left.right.right = new TreeNode(9);
        tree.root.right.left.right = new TreeNode(12);
        /*
                  1          
                 /  \
              → 11   3       
               /    /  \
              4    5    6    
             / \    \    \   
            8   7    12   -4  
               / \       
             -2   9      
            --------------------
            Max left node is : 11
        */
        tree.maxLeftChild();
    }
}

Output

11
<?php
// Php program for
// Find maximum among all left nodes in Binary Tree
class TreeNode
{
    // Data value
    public $data;
    // Indicates left and right subtree
    public $left;
    public $right;
    public  function __construct($data)
    {
        $this->data = $data;
        $this->left = NULL;
        $this->right = NULL;
    }
}
class BinaryTree
{
    public $root;
    public $result;
    public  function __construct()
    {
        $this->root = NULL;
        $this->result = NULL;
    }
    public  function findMaxLeftNode($node)
    {
        if ($node != NULL)
        {
            if ($node->left != NULL)
            {
                if ($this->result == NULL || 
                    $this->result->data < $node->left->data)
                {
                    // Get new max left child node
                    $this->result = $node->left;
                }
            }
            // Visit left subtree
            $this->findMaxLeftNode($node->left);
            // Visit right subtree
            $this->findMaxLeftNode($node->right);
        }
    }
    public  function maxLeftChild()
    {
        $this->result = NULL;
        $this->findMaxLeftNode($this->root);
        if ($this->result == NULL)
        {
            // When no left child node
            echo("None\n");
        }
        else
        {
            // Display calculated result
            echo($this->result->data.
                "\n");
        }
    }
}

function main()
{
    $tree = new BinaryTree();
    /*
        Binary Tree
        ---------------
              1
             /  \
            11   3
           /    /  \
          4    5    6   
         / \    \    \  
        8   7    12   -4
           / \       
         -2   9    
    */
    // Add Binary tree nodes
    $tree->root = new TreeNode(1);
    $tree->root->left = new TreeNode(11);
    $tree->root->right = new TreeNode(3);
    $tree->root->right->right = new TreeNode(6);
    $tree->root->right->right->right = new TreeNode(-4);
    $tree->root->right->left = new TreeNode(5);
    $tree->root->left->left = new TreeNode(4);
    $tree->root->left->left->left = new TreeNode(8);
    $tree->root->left->left->right = new TreeNode(7);
    $tree->root->left->left->right->left = new TreeNode(-2);
    $tree->root->left->left->right->right = new TreeNode(9);
    $tree->root->right->left->right = new TreeNode(12);
    /*
              1          
             /  \
          → 11   3       
           /    /  \
          4    5    6    
         / \    \    \   
        8   7    12   -4  
           / \       
         -2   9      
        --------------------
        Max left node is : 11
    */
    $tree->maxLeftChild();
}
main();

Output

11
// Node JS program for
// Find maximum among all left nodes in Binary Tree
class TreeNode
{
    constructor(data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
class BinaryTree
{
    constructor()
    {
        this.root = null;
        this.result = null;
    }
    findMaxLeftNode(node)
    {
        if (node != null)
        {
            if (node.left != null)
            {
                if (this.result == null || 
                    this.result.data < node.left.data)
                {
                    // Get new max left child node
                    this.result = node.left;
                }
            }
            // Visit left subtree
            this.findMaxLeftNode(node.left);
            // Visit right subtree
            this.findMaxLeftNode(node.right);
        }
    }
    maxLeftChild()
    {
        this.result = null;
        this.findMaxLeftNode(this.root);
        if (this.result == null)
        {
            // When no left child node
            console.log("None");
        }
        else
        {
            // Display calculated result
            console.log(this.result.data);
        }
    }
}

function main()
{
    var tree = new BinaryTree();
    /*
        Binary Tree
        ---------------
              1
             /  \
            11   3
           /    /  \
          4    5    6   
         / \    \    \  
        8   7    12   -4
           / \       
         -2   9    
    */
    // Add Binary tree nodes
    tree.root = new TreeNode(1);
    tree.root.left = new TreeNode(11);
    tree.root.right = new TreeNode(3);
    tree.root.right.right = new TreeNode(6);
    tree.root.right.right.right = new TreeNode(-4);
    tree.root.right.left = new TreeNode(5);
    tree.root.left.left = new TreeNode(4);
    tree.root.left.left.left = new TreeNode(8);
    tree.root.left.left.right = new TreeNode(7);
    tree.root.left.left.right.left = new TreeNode(-2);
    tree.root.left.left.right.right = new TreeNode(9);
    tree.root.right.left.right = new TreeNode(12);
    /*
              1          
             /  \
          → 11   3       
           /    /  \
          4    5    6    
         / \    \    \   
        8   7    12   -4  
           / \       
         -2   9      
        --------------------
        Max left node is : 11
    */
    tree.maxLeftChild();
}
main();

Output

11
#  Python 3 program for
#  Find maximum among all left nodes in Binary Tree
class TreeNode :
    #  Data value
    #  Indicates left and right subtree
    def __init__(self, data) :
        self.data = data
        self.left = None
        self.right = None
    

class BinaryTree :
    def __init__(self) :
        self.root = None
        self.result = None
    
    def findMaxLeftNode(self, node) :
        if (node != None) :
            if (node.left != None) :
                if (self.result == None or 
                    self.result.data < node.left.data) :
                    #  Get new max left child node
                    self.result = node.left
                
            
            #  Visit left subtree
            self.findMaxLeftNode(node.left)
            #  Visit right subtree
            self.findMaxLeftNode(node.right)
        
    
    def maxLeftChild(self) :
        self.result = None
        self.findMaxLeftNode(self.root)
        if (self.result == None) :
            #  When no left child node
            print("None")
        else :
            #  Display calculated result
            print(self.result.data)
        
    

def main() :
    tree = BinaryTree()
    #    Binary Tree
    #    ---------------
    #          1
    #         /  \
    #        11   3
    #       /    /  \
    #      4    5    6   
    #     / \    \    \  
    #    8   7    12   -4
    #       / \       
    #     -2   9    
    #  Add Binary tree nodes
    tree.root = TreeNode(1)
    tree.root.left = TreeNode(11)
    tree.root.right = TreeNode(3)
    tree.root.right.right = TreeNode(6)
    tree.root.right.right.right = TreeNode(-4)
    tree.root.right.left = TreeNode(5)
    tree.root.left.left = TreeNode(4)
    tree.root.left.left.left = TreeNode(8)
    tree.root.left.left.right = TreeNode(7)
    tree.root.left.left.right.left = TreeNode(-2)
    tree.root.left.left.right.right = TreeNode(9)
    tree.root.right.left.right = TreeNode(12)
    #          1          
    #         /  \
    #      → 11   3       
    #       /    /  \
    #      4    5    6    
    #     / \    \    \   
    #    8   7    12   -4  
    #       / \       
    #     -2   9      
    #    --------------------
    #    Max left node is : 11
    tree.maxLeftChild()

if __name__ == "__main__": main()

Output

11
#  Ruby program for
#  Find maximum among all left nodes in Binary Tree
class TreeNode 
    # Define the accessor and reader of class TreeNode
    attr_reader :data, :left, :right
    attr_accessor :data, :left, :right
    #  Data value
    #  Indicates left and right subtree
    def initialize(data) 
        self.data = data
        self.left = nil
        self.right = nil
    end

end

class BinaryTree 
    # Define the accessor and reader of class BinaryTree
    attr_reader :root, :result
    attr_accessor :root, :result
    def initialize() 
        self.root = nil
        self.result = nil
    end

    def findMaxLeftNode(node) 
        if (node != nil) 
            if (node.left != nil) 
                if (self.result == nil || 
                    self.result.data < node.left.data) 
                    #  Get new max left child node
                    self.result = node.left
                end

            end

            #  Visit left subtree
            self.findMaxLeftNode(node.left)
            #  Visit right subtree
            self.findMaxLeftNode(node.right)
        end

    end

    def maxLeftChild() 
        self.result = nil
        self.findMaxLeftNode(self.root)
        if (self.result == nil) 
            #  When no left child node
            print("None", "\n")
        else
 
            #  Display calculated result
            print(self.result.data, "\n")
        end

    end

end

def main() 
    tree = BinaryTree.new()
    #    Binary Tree
    #    ---------------
    #          1
    #         /  \
    #        11   3
    #       /    /  \
    #      4    5    6   
    #     / \    \    \  
    #    8   7    12   -4
    #       / \       
    #     -2   9    
    #  Add Binary tree nodes
    tree.root = TreeNode.new(1)
    tree.root.left = TreeNode.new(11)
    tree.root.right = TreeNode.new(3)
    tree.root.right.right = TreeNode.new(6)
    tree.root.right.right.right = TreeNode.new(-4)
    tree.root.right.left = TreeNode.new(5)
    tree.root.left.left = TreeNode.new(4)
    tree.root.left.left.left = TreeNode.new(8)
    tree.root.left.left.right = TreeNode.new(7)
    tree.root.left.left.right.left = TreeNode.new(-2)
    tree.root.left.left.right.right = TreeNode.new(9)
    tree.root.right.left.right = TreeNode.new(12)
    #          1          
    #         /  \
    #      → 11   3       
    #       /    /  \
    #      4    5    6    
    #     / \    \    \   
    #    8   7    12   -4  
    #       / \       
    #     -2   9      
    #    --------------------
    #    Max left node is : 11
    tree.maxLeftChild()
end

main()

Output

11

// Scala program for
// Find maximum among all left nodes in Binary Tree
class TreeNode(
    // Data value
    var data: Int,
        // Indicates left and right subtree
        var left: TreeNode,
            var right: TreeNode)
{
    def this(data: Int)
    {
        this(data, null, null);
    }
}
class BinaryTree(var root: TreeNode,
    var result: TreeNode)
{
    def this()
    {
        this(null, null);
    }
    def findMaxLeftNode(node: TreeNode): Unit = {
        if (node != null)
        {
            if (node.left != null)
            {
                if (this.result == null || 
                    this.result.data < node.left.data)
                {
                    // Get new max left child node
                    this.result = node.left;
                }
            }
            // Visit left subtree
            findMaxLeftNode(node.left);
            // Visit right subtree
            findMaxLeftNode(node.right);
        }
    }
    def maxLeftChild(): Unit = {
        this.result = null;
        this.findMaxLeftNode(this.root);
        if (this.result == null)
        {
            // When no left child node
            println("None");
        }
        else
        {
            // Display calculated result
            println(this.result.data);
        }
    }
}
object Main
{
    def main(args: Array[String]): Unit = {
        var tree: BinaryTree = new BinaryTree();
        /*
            Binary Tree
            ---------------
                  1
                 /  \
                11   3
               /    /  \
              4    5    6   
             / \    \    \  
            8   7    12   -4
               / \       
             -2   9    
        */
        // Add Binary tree nodes
        tree.root = new TreeNode(1);
        tree.root.left = new TreeNode(11);
        tree.root.right = new TreeNode(3);
        tree.root.right.right = new TreeNode(6);
        tree.root.right.right.right = new TreeNode(-4);
        tree.root.right.left = new TreeNode(5);
        tree.root.left.left = new TreeNode(4);
        tree.root.left.left.left = new TreeNode(8);
        tree.root.left.left.right = new TreeNode(7);
        tree.root.left.left.right.left = new TreeNode(-2);
        tree.root.left.left.right.right = new TreeNode(9);
        tree.root.right.left.right = new TreeNode(12);
        /*
                  1          
                 /  \
              → 11   3       
               /    /  \
              4    5    6    
             / \    \    \   
            8   7    12   -4  
               / \       
             -2   9      
            --------------------
            Max left node is : 11
        */
        tree.maxLeftChild();
    }
}

Output

11
// Swift 4 program for
// Find maximum among all left nodes in Binary Tree
class TreeNode
{
    // Data value
    var data: Int;
    // Indicates left and right subtree
    var left: TreeNode? ;
    var right: TreeNode? ;
    init(_ data: Int)
    {
        self.data = data;
        self.left = nil;
        self.right = nil;
    }
}
class BinaryTree
{
    var root: TreeNode? ;
    var result: TreeNode? ;
    init()
    {
        self.root = nil;
        self.result = nil;
    }
    func findMaxLeftNode(_ node: TreeNode? )
    {
        if (node  != nil)
        {
            if (node!.left  != nil)
            {
                if (self.result == nil || 
                    self.result!.data < node!.left!.data)
                {
                    // Get new max left child node
                    self.result = node!.left;
                }
            }
            // Visit left subtree
            self.findMaxLeftNode(node!.left);
            // Visit right subtree
            self.findMaxLeftNode(node!.right);
        }
    }
    func maxLeftChild()
    {
        self.result = nil;
        self.findMaxLeftNode(self.root);
        if (self.result == nil)
        {
            // When no left child node
            print("None");
        }
        else
        {
            // Display calculated result
            print(self.result!.data);
        }
    }
}
func main()
{
    let tree: BinaryTree = BinaryTree();
    /*
        Binary Tree
        ---------------
              1
             /  \
            11   3
           /    /  \
          4    5    6   
         / \    \    \  
        8   7    12   -4
           / \       
         -2   9    
    */
    // Add Binary tree nodes
    tree.root = TreeNode(1);
    tree.root!.left = TreeNode(11);
    tree.root!.right = TreeNode(3);
    tree.root!.right!.right = TreeNode(6);
    tree.root!.right!.right!.right = TreeNode(-4);
    tree.root!.right!.left = TreeNode(5);
    tree.root!.left!.left = TreeNode(4);
    tree.root!.left!.left!.left = TreeNode(8);
    tree.root!.left!.left!.right = TreeNode(7);
    tree.root!.left!.left!.right!.left = TreeNode(-2);
    tree.root!.left!.left!.right!.right = TreeNode(9);
    tree.root!.right!.left!.right = TreeNode(12);
    /*
              1          
             /  \
          → 11   3       
           /    /  \
          4    5    6    
         / \    \    \   
        8   7    12   -4  
           / \       
         -2   9      
        --------------------
        Max left node is : 11
    */
    tree.maxLeftChild();
}
main();

Output

11
// Kotlin program for
// Find maximum among all left nodes in Binary Tree
class TreeNode
{
    // Data value
    var data: Int;
    // Indicates left and right subtree
    var left: TreeNode ? ;
    var right: TreeNode ? ;
    constructor(data: Int)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
class BinaryTree
{
    var root: TreeNode ? ;
    var result: TreeNode ? ;
    constructor()
    {
        this.root = null;
        this.result = null;
    }
    fun findMaxLeftNode(node: TreeNode ? ): Unit
    {
        if (node != null)
        {
            if (node.left != null)
            {
                if (this.result == null || 
                    this.result!!.data < node.left!!.data)
                {
                    // Get new max left child node
                    this.result = node.left;
                }
            }
            // Visit left subtree
            this.findMaxLeftNode(node.left);
            // Visit right subtree
            this.findMaxLeftNode(node.right);
        }
    }
    fun maxLeftChild(): Unit
    {
        this.result = null;
        this.findMaxLeftNode(this.root);
        if (this.result == null)
        {
            // When no left child node
            println("None");
        }
        else
        {
            // Display calculated result
            println(this.result?.data);
        }
    }
}
fun main(args: Array < String > ): Unit
{
    val tree: BinaryTree = BinaryTree();
    /*
        Binary Tree
        ---------------
              1
             /  \
            11   3
           /    /  \
          4    5    6   
         / \    \    \  
        8   7    12   -4
           / \       
         -2   9    
    */
    // Add Binary tree nodes
    tree.root = TreeNode(1);
    tree.root?.left = TreeNode(11);
    tree.root?.right = TreeNode(3);
    tree.root?.right?.right = TreeNode(6);
    tree.root?.right?.right?.right = TreeNode(-4);
    tree.root?.right?.left = TreeNode(5);
    tree.root?.left?.left = TreeNode(4);
    tree.root?.left?.left?.left = TreeNode(8);
    tree.root?.left?.left?.right = TreeNode(7);
    tree.root?.left?.left?.right?.left = TreeNode(-2);
    tree.root?.left?.left?.right?.right = TreeNode(9);
    tree.root?.right?.left?.right = TreeNode(12);
    /*
              1          
             /  \
          → 11   3       
           /    /  \
          4    5    6    
         / \    \    \   
        8   7    12   -4  
           / \       
         -2   9      
        --------------------
        Max left node is : 11
    */
    tree.maxLeftChild();
}

Output

11

The time complexity in the above program is O(n), And this problem of solved by using of recursion. Your task is to solve this problem using iteration.



Please share your knowledge to improve code and content standard. Also submit your doubts, and test case. We improve by your feedback. We will try to resolve your query as soon as possible.

New Comment




© 2021, kalkicode.com, All rights reserved