Deepest left leaf node in a binary tree | iterative approach

Given a Binary Tree, find the deepest leaf node that is left child of its parent. For example, consider the following tree. The deepest left leaf node is the node with value 9.

Examples: 

Input : 
       1
     /   \
    2     3
  /      /  \  
 4      5    6
        \     \
         7     8
        /       \
       9         10


Output : 9

Recursive approach to this problem is discussed here
For iterative approach, idea is similar to Method 2 of level order traversal
The idea is to traverse the tree iteratively and whenever a left tree node is pushed to queue, check if it is leaf node, if it's leaf node, then update the result. Since we go level by level, the last stored leaf node is deepest one, 

Algorithm:

• Define a Node structure that has data, left, and right pointers.
• Create a function newNode that takes data as an argument and returns a new node with data set to data, and left and right pointers set to NULL.
• Create a function getDeepestLeftLeafNode that takes the root of a binary tree as an argument and returns the deepest left leaf node of the tree.
  • Initialize a queue q and add the root node of the binary tree to the queue.
  • Initialize a pointer result to NULL.
  • Use a while loop to traverse the binary tree level by level until the queue becomes empty.
    • Within the while loop, pop the front element of the queue and store it in a temporary pointer temp.
    • If temp has a left child, add it to the queue, and check if it is a leaf node (i.e., it does not have left or right child).
    • If the left child of temp is a leaf node, set result to temp->left.
    • If temp has a right child, add it to the queue.
  • Return result which will contain the deepest left leaf node of the binary tree.
• In the main function, construct a binary tree and call getDeepestLeftLeafNode to find the deepest left leaf node of the tree.
  • If result is not NULL, print the data of the deepest left leaf node, otherwise print "No result, left leaf not found".

Implementation:

// CPP program to find deepest left leaf
// node of binary tree
#include <bits/stdc++.h>
using namespace std;

// tree node
struct Node {
    int data;
    Node *left, *right;
};

// returns a new tree Node
Node* newNode(int data)
{
    Node* temp = new Node();
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}

// return the deepest left leaf node
// of binary tree
Node* getDeepestLeftLeafNode(Node* root)
{
    if (!root)
        return NULL;

    // create a queue for level order traversal
    queue<Node*> q;
    q.push(root);

    Node* result = NULL;

    // traverse until the queue is empty
    while (!q.empty()) {
        Node* temp = q.front();
        q.pop();

         
        // Since we go level by level, the last 
        // stored left leaf node is deepest one,
        if (temp->left) {
            q.push(temp->left);
            if (!temp->left->left && !temp->left->right)
                result = temp->left;
        }
        
        if (temp->right)
            q.push(temp->right);
    }
    return result;
}

// driver program
int main()
{
    // construct a tree
    Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->right->left = newNode(5);
    root->right->right = newNode(6);
    root->right->left->right = newNode(7);
    root->right->right->right = newNode(8);
    root->right->left->right->left = newNode(9);
    root->right->right->right->right = newNode(10);

    Node* result = getDeepestLeftLeafNode(root);
    if (result)
        cout << "Deepest Left Leaf Node :: " 
             << result->data << endl;
    else
        cout << "No result, left leaf not found\n";
    return 0;
}

// Java program to find deepest left leaf
// node of binary tree
import java.util.*;

public class GFG 
{

// tree node
static class Node
{
    int data;
    Node left, right;
};

// returns a new tree Node
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
}

// return the deepest left leaf node
// of binary tree
static Node getDeepestLeftLeafNode(Node root)
{
    if (root == null)
        return null;

    // create a queue for level order traversal
    Queue<Node> q = new LinkedList<>();
    q.add(root);

    Node result = null;

    // traverse until the queue is empty
    while (!q.isEmpty()) 
    {
        Node temp = q.peek();
        q.remove();

        // Since we go level by level, the last 
        // stored left leaf node is deepest one,
        if (temp.left != null) 
        {
            q.add(temp.left);
            if (temp.left.left == null &&
                temp.left.right == null)
                result = temp.left;
        }
        
        if (temp.right != null)
            q.add(temp.right);
    }
    return result;
}

// Driver Code
public static void main(String[] args)
{
    
    // construct a tree
    Node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.right.left = newNode(5);
    root.right.right = newNode(6);
    root.right.left.right = newNode(7);
    root.right.right.right = newNode(8);
    root.right.left.right.left = newNode(9);
    root.right.right.right.right = newNode(10);

    Node result = getDeepestLeftLeafNode(root);
    if (result != null)
        System.out.println("Deepest Left Leaf Node :: " + 
                                            result.data);
    else
        System.out.println("No result, " + 
                   "left leaf not found");
    }
}

// This code is contributed by Rajput-Ji

# Python3 program to find deepest 
# left leaf Binary search Tree

_MIN = -2147483648
_MAX = 2147483648

# Helper function that allocates a new 
# node with the given data and None 
# left and right pointers.                                     
class newnode: 

    # Constructor to create a new node 
    def __init__(self, data): 
        self.data = data 
        self.left = None
        self.right = None

# utility function to return deepest
# left leaf node
def getDeepestLeftLeafNode(root) :

    if (not root):
        return None

    # create a queue for level
    # order traversal 
    q = [] 
    q.append(root) 

    result = None

    # traverse until the queue is empty 
    while (len(q)): 
        temp = q[0] 
        q.pop(0) 

        if (temp.left):
            q.append(temp.left)
            if (not temp.left.left and 
                not temp.left.right): 
                result = temp.left 
        
        # Since we go level by level, 
        # the last stored right leaf
        # node is deepest one 
        if (temp.right): 
            q.append(temp.right)         
    
    return result

# Driver Code 
if __name__ == '__main__':
    
    # create a binary tree 
    root = newnode(1) 
    root.left = newnode(2) 
    root.right = newnode(3) 
    root.left.Left = newnode(4)
    root.right.left = newnode(5) 
    root.right.right = newnode(6)
    root.right.left.right = newnode(7)
    root.right.right.right = newnode(8)
    root.right.left.right.left = newnode(9)
    root.right.right.right.right = newnode(10) 

    result = getDeepestLeftLeafNode(root)
    if result:
        print("Deepest Left Leaf Node ::",
                              result.data)
    else:
        print("No result, Left leaf not found")
        
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

// C# program to find deepest left leaf
// node of binary tree
using System;
using System.Collections.Generic;
    
class GFG 
{

// tree node
class Node
{
    public int data;
    public Node left, right;
};

// returns a new tree Node
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
}

// return the deepest left leaf node
// of binary tree
static Node getDeepestLeftLeafNode(Node root)
{
    if (root == null)
        return null;

    // create a queue for level order traversal
    Queue<Node> q = new Queue<Node>();
    q.Enqueue(root);

    Node result = null;

    // traverse until the queue is empty
    while (q.Count != 0) 
    {
        Node temp = q.Peek();
        q.Dequeue();

        // Since we go level by level, the last 
        // stored left leaf node is deepest one,
        if (temp.left != null) 
        {
            q.Enqueue(temp.left);
            if (temp.left.left == null &&
                temp.left.right == null)
                result = temp.left;
        }
        if (temp.right != null)
            q.Enqueue(temp.right);
    }
    return result;
}

// Driver Code
public static void Main(String[] args)
{
    
    // construct a tree
    Node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.right.left = newNode(5);
    root.right.right = newNode(6);
    root.right.left.right = newNode(7);
    root.right.right.right = newNode(8);
    root.right.left.right.left = newNode(9);
    root.right.right.right.right = newNode(10);

    Node result = getDeepestLeftLeafNode(root);
    if (result != null)
        Console.WriteLine("Deepest Left Leaf Node :: " + 
                                           result.data);
    else
        Console.WriteLine("No result, " + 
                  "left leaf not found");
    }
}

// This code is contributed by Rajput-Ji

<script>

    // JavaScript program to find deepest
    // left leaf node of binary tree
    
    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }
    
    // returns a new tree Node
    function newNode(data)
    {
        let temp = new Node(data);
        return temp;
    }

    // return the deepest left leaf node
    // of binary tree
    function getDeepestLeftLeafNode(root)
    {
        if (root == null)
            return null;

        // create a queue for level order traversal
        let q = [];
        q.push(root);

        let result = null;

        // traverse until the queue is empty
        while (q.length > 0) 
        {
            let temp = q[0];
            q.shift();

            // Since we go level by level, the last 
            // stored left leaf node is deepest one,
            if (temp.left != null) 
            {
                q.push(temp.left);
                if (temp.left.left == null &&
                    temp.left.right == null)
                    result = temp.left;
            }

            if (temp.right != null)
                q.push(temp.right);
        }
        return result;
    }
    
    // construct a tree
    let root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.left = newNode(4);
    root.right.left = newNode(5);
    root.right.right = newNode(6);
    root.right.left.right = newNode(7);
    root.right.right.right = newNode(8);
    root.right.left.right.left = newNode(9);
    root.right.right.right.right = newNode(10);
  
    let result = getDeepestLeftLeafNode(root);
    if (result != null)
        document.write("Deepest Left Leaf Node :: " + 
                                            result.data);
    else
        document.write("No result, " + 
                   "left leaf not found");

</script>

Deepest Left Leaf Node :: 9

Time complexity: O(n) where n is no of nodes in a given binary tree
Auxiliary space: O(n)