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Deepest left leaf node in a binary tree

Last Updated : 07 Mar, 2023
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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. 

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

The idea is to recursively traverse the given binary tree and while traversing, maintain “level” which will store the current node’s level in the tree. If current node is left leaf, then check if its level is more than the level of deepest left leaf seen so far. If level is more then update the result. If current node is not leaf, then recursively find maximum depth in left and right subtrees, and return maximum of the two depths.

Implementation:

C++ 
// A C++ program to find the deepest left leaf in a given binary tree
#include <bits/stdc++.h>
using namespace std;

struct Node
{
    int val;
    struct Node *left, *right;
};

Node *newNode(int data)
{
    Node *temp = new Node;
    temp->val = data;
    temp->left = temp->right =  NULL;
    return temp;
}

// A utility function to find deepest leaf node.
// lvl:  level of current node.
// maxlvl: pointer to the deepest left leaf node found so far
// isLeft: A bool indicate that this node is left child of its parent
// resPtr: Pointer to the result
void deepestLeftLeafUtil(Node *root, int lvl, int *maxlvl,
                         bool isLeft, Node **resPtr)
{
    // Base case
    if (root == NULL)
        return;

    // Update result if this node is left leaf and its level is more
    // than the maxl level of the current result
    if (isLeft && !root->left && !root->right && lvl > *maxlvl)
    {
        *resPtr = root;
        *maxlvl = lvl;
        return;
    }

    // Recur for left and right subtrees
    deepestLeftLeafUtil(root->left, lvl+1, maxlvl, true, resPtr);
    deepestLeftLeafUtil(root->right, lvl+1, maxlvl, false, resPtr);
}

// A wrapper over deepestLeftLeafUtil().
Node* deepestLeftLeaf(Node *root)
{
    int maxlevel = 0;
    Node *result = NULL;
    deepestLeftLeafUtil(root, 0, &maxlevel, false, &result);
    return result;
}

// Driver program to test above function
int main()
{
    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 = deepestLeftLeaf(root);
    if (result)
        cout << "The deepest left child is " << result->val;
    else
        cout << "There is no left leaf in the given tree";

    return 0;
}
Java 
// A Java program to find
// the deepest left leaf
// in a binary tree

// A Binary Tree node
class Node
{
    int data;
    Node left, right;

    // Constructor
    public Node(int data)
    {
        this.data = data;
        left = right = null;
    }
}

// Class to evaluate pass
// by reference 
class Level 
{
    // maxlevel: gives the
    // value of level of
    // maximum left leaf
    int maxlevel = 0;
}

class BinaryTree 
{
    Node root;
    
    // Node to store resultant
    // node after left traversal
    Node result;

    // A utility function to
    // find deepest leaf node.
    // lvl: level of current node.
    // isLeft: A bool indicate
    // that this node is left child
    void deepestLeftLeafUtil(Node node, 
                             int lvl, 
                             Level level,
                             boolean isLeft) 
    {
        // Base case
        if (node == null) 
            return;

        // Update result if this node
        // is left leaf and its level
        // is more than the maxl level
        // of the current result
        if (isLeft != false &&
            node.left == null &&
            node.right == null &&
            lvl > level.maxlevel)
        {
            result = node;
            level.maxlevel = lvl;
        }

        // Recur for left and right subtrees
        deepestLeftLeafUtil(node.left, lvl + 1,
                            level, true);
        deepestLeftLeafUtil(node.right, lvl + 1,
                            level, false);
    }

    // A wrapper over deepestLeftLeafUtil().
    void deepestLeftLeaf(Node node) 
    {
        Level level = new Level();
        deepestLeftLeafUtil(node, 0, level, false);
    }
    
    // Driver program to test above functions
    public static void main(String[] args) 
    {
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.left = new Node(4);
        tree.root.right.left = new Node(5);
        tree.root.right.right = new Node(6);
        tree.root.right.left.right = new Node(7);
        tree.root.right.right.right = new Node(8);
        tree.root.right.left.right.left = new Node(9);
        tree.root.right.right.right.right = new Node(10);

        tree.deepestLeftLeaf(tree.root);
        if (tree.result != null)
            System.out.println("The deepest left child"+
                               " is " + tree.result.data);
        else
            System.out.println("There is no left leaf in"+
                               " the given tree");
    }
}

// This code has been contributed by Mayank Jaiswal(mayank_24)
Python3 
# Python program to find the deepest left leaf in a given
# Binary tree

# A binary tree node
class Node:
    
    # Constructor to create a new node
    def __init__(self, val):
        self.val = val 
        self.left = None
        self.right = None

# A utility function to find deepest leaf node.
# lvl:  level of current node.
# maxlvl: pointer to the deepest left leaf node found so far
# isLeft: A bool indicate that this node is left child
# of its parent
# resPtr: Pointer to the result
def deepestLeftLeafUtil(root, lvl, maxlvl, isLeft):
    
    # Base CAse
    if root is None:
        return

    # Update result if this node is left leaf and its 
    # level is more than the max level of the current result
    if(isLeft is True):
        if (root.left == None and root.right == None):
            if lvl > maxlvl[0] : 
                deepestLeftLeafUtil.resPtr = root 
                maxlvl[0] = lvl 
                return

    # Recur for left and right subtrees
    deepestLeftLeafUtil(root.left, lvl+1, maxlvl, True)
    deepestLeftLeafUtil(root.right, lvl+1, maxlvl, False)

# A wrapper for left and right subtree
def deepestLeftLeaf(root):
    maxlvl = [0]
    deepestLeftLeafUtil.resPtr = None
    deepestLeftLeafUtil(root, 0, maxlvl, False)
    return deepestLeftLeafUtil.resPtr


# Driver program to test above function
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.right.left = Node(5)
root.right.right = Node(6)
root.right.left.right = Node(7)
root.right.right.right = Node(8)
root.right.left.right.left = Node(9)
root.right.right.right.right= Node(10)

result = deepestLeftLeaf(root) 

if result is None:
    print ("There is not left leaf in the given tree")
else:
    print ("The deepst left child is", result.val)

# This code is contributed by Nikhil Kumar Singh(nickzuck_007)
C# 
using System;

// A C# program to find 
// the deepest left leaf 
// in a binary tree 

// A Binary Tree node 
public class Node
{
    public int data;
    public Node left, right;

    // Constructor 
    public Node(int data)
    {
        this.data = data;
        left = right = null;
    }
}

// Class to evaluate pass 
// by reference 
public class Level
{
    // maxlevel: gives the 
    // value of level of 
    // maximum left leaf 
    public int maxlevel = 0;
}

public class BinaryTree
{
    public Node root;

    // Node to store resultant 
    // node after left traversal 
    public Node result;

    // A utility function to 
    // find deepest leaf node. 
    // lvl: level of current node. 
    // isLeft: A bool indicate 
    // that this node is left child 
    public virtual void deepestLeftLeafUtil(Node node, int lvl, 
                                        Level level, bool isLeft)
    {
        // Base case 
        if (node == null)
        {
            return;
        }

        // Update result if this node 
        // is left leaf and its level 
        // is more than the maxl level 
        // of the current result 
        if (isLeft != false && node.left == null && node.right == null
                            && lvl > level.maxlevel)
        {
            result = node;
            level.maxlevel = lvl;
        }

        // Recur for left and right subtrees 
        deepestLeftLeafUtil(node.left, lvl + 1, level, true);
        deepestLeftLeafUtil(node.right, lvl + 1, level, false);
    }

    // A wrapper over deepestLeftLeafUtil(). 
    public virtual void deepestLeftLeaf(Node node)
    {
        Level level = new Level();
        deepestLeftLeafUtil(node, 0, level, false);
    }

    // Driver program to test above functions 
    public static void Main(string[] args)
    {
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.left = new Node(4);
        tree.root.right.left = new Node(5);
        tree.root.right.right = new Node(6);
        tree.root.right.left.right = new Node(7);
        tree.root.right.right.right = new Node(8);
        tree.root.right.left.right.left = new Node(9);
        tree.root.right.right.right.right = new Node(10);

        tree.deepestLeftLeaf(tree.root);
        if (tree.result != null)
        {
            Console.WriteLine("The deepest left child is " + tree.result.data);
        }
        else
        {
            Console.WriteLine("There is no left leaf in the given tree");
        }
    }
}

// This code is contributed by Shrikant13
JavaScript 
<script>

// A Javascript program to find
// the deepest left leaf
// in a binary tree
class Node
{
    constructor(data)
    {
        this.left = null;
        this.right = null;
        this.data = data;
    }
}

let maxlevel = 0;
let root;
  
// Node to store resultant
// node after left traversal
let result;

// A utility function to
// find deepest leaf node.
// lvl: level of current node.
// isLeft: A bool indicate
// that this node is left child
function deepestLeftLeafUtil(node, lvl, isLeft) 
{
    
    // Base case
    if (node == null) 
        return;

    // Update result if this node
    // is left leaf and its level
    // is more than the maxl level
    // of the current result
    if (isLeft != false &&
        node.left == null &&
        node.right == null &&
        lvl > maxlevel)
    {
        result = node;
        maxlevel = lvl;
    }

    // Recur for left and right subtrees
    deepestLeftLeafUtil(node.left, lvl + 1, true);
    deepestLeftLeafUtil(node.right, lvl + 1, false);
}

// A wrapper over deepestLeftLeafUtil().
function deepestLeftLeaf(node) 
{
    deepestLeftLeafUtil(node, 0, false);
}

// Driver code
root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.right.left = new Node(5);
root.right.right = new Node(6);
root.right.left.right = new Node(7);
root.right.right.right = new Node(8);
root.right.left.right.left = new Node(9);
root.right.right.right.right = new Node(10);

deepestLeftLeaf(root);
if (result != null)
  document.write("The deepest left child"+
                 " is " + result.data + "</br>");
else
  document.write("There is no left leaf in"+
                 " the given tree");
                 
// This code is contributed by decode2207

</script>

Output
The deepest left child is 9

Time Complexity: The function does a simple traversal of the tree, so the complexity is O(n). 

Auxiliary Space: O(n) for call stack because using recursion

Iterative Approach(Using Level Order Traversal):
Follow the below steps to solve the above problem:
1) We simply traverse the tree using Level Order Traversal with queue data structure.
2) If current node has left child then we update our answer with left child.
3) Finally return the ans node.

Below is the implementation of above approach:

C++ 
// Iterative Approach to solve this problem
// C++ code for above approach
// This code is contributed by Yash Agarwal(yashagarwal2852002)
#include<bits/stdc++.h>
using namespace std;

// a binary tree node
struct Node{
    int data;
    Node* left;
    Node* right;
    Node(int data){
        this->data = data;
        this->left = NULL;
        this->right = NULL;
    }
};

// utility function to create a new tree node
Node* newNode(int data){
    return new Node(data);
}

// function will return the deepest left leaf node
Node* deepestLeftLeaf(Node* root){
    if(root == NULL) return NULL;
    Node* ans = NULL;
    queue<Node*> q;
    q.push(root);
    while(!q.empty()){
        Node* front_node = q.front();
        q.pop();
        if(front_node->left){
            q.push(front_node->left);
            ans = front_node->left;
        }
        if(front_node->right){
            q.push(front_node->right);
        }
    }
    return ans;
}

// Driver program to test above function
int main()
{
    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 = deepestLeftLeaf(root);
    if (result)
        cout << "The deepest left child is " << result->data;
    else
        cout << "There is no left leaf in the given tree";
 
    return 0;
}
// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002)
Java 
import java.util.LinkedList;
import java.util.Queue;

class Node {
    int data;
    Node left, right;

    public Node(int item) {
        data = item;
        left = right = null;
    }
}

public class Main {
    // utility function create a new node
    static Node newNode(int data) {
        return new Node(data);
    }

    // function will return the deepest left leaf node
    static Node deepestLeftLeaf(Node root) {
        if (root == null) return null;
        Node ans = null;
        Queue<Node> q = new LinkedList<Node>();
        q.add(root);
        while (!q.isEmpty()) {
            Node front_node = q.poll();
            if (front_node.left != null) {
                q.add(front_node.left);
                ans = front_node.left;
            }
            if (front_node.right != null) q.add(front_node.right);
        }
        return ans;
    }

    // driver code to test above function
    public static void main(String[] args) {
        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 = deepestLeftLeaf(root);

        if (result != null) {
            System.out.println("The deepest left child is : " + result.data);
        } else {
            System.out.println("There is no left leaf in the given tree.");
        }
    }
}
Python 
# Iterative approach to solve this problem
# Python code for the above approach
# a binary tree node
class Node:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
    
# utility function to create a new tree node
def newNode(data):
    return Node(data)

# // function will return the deepest left leaf node
def deepestLeftLeaf(root):
    if(root is None):
        return
    ans = None
    q = []
    q.append(root)
    while(len(q) > 0):
        front_node = q.pop(0)
        if(front_node.left is not None):
            q.append(front_node.left)
            ans = front_node.left
        if(front_node.right is not None):
            q.append(front_node.right)
        
    return ans

# driver program to test above function
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 = deepestLeftLeaf(root)
if(result is not None):
    print("The deepest left child is : ")
    print(result.data)
else:
    print("There is no left leaf in the given tree.")
C# 
// C# Program to find sum of all right leaves
using System;
using System.Collections.Generic;

public class Node{
  public int data;
  public Node left, right;
  public Node(int item){
    data = item;
    left = right = null;
  }
}

class GFG{
  // utility function create a new node
  static Node newNode(int data){
    return new Node(data);
  }

  // function will return the deepest left leaf node
  static Node deepestLeftLeaf(Node root){
    if(root == null) return null;
    Node ans = null;
    Queue<Node> q = new Queue<Node>();
    q.Enqueue(root);
    while(q.Count > 0){
      Node front_node = q.Dequeue();
      if(front_node.left != null){
        q.Enqueue(front_node.left);
        ans = front_node.left;
      }
      if(front_node.right != null) q.Enqueue(front_node.right);
    }
    return ans;
  }

  // driver code to test above function
  public static void Main(String[] args){
    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 = deepestLeftLeaf(root);

    if(result != null){
      Console.WriteLine("The deepest left child is : " + result.data);
    }
    else{
      Console.WriteLine("There is no left leaf in the given tree.");
    }
  }
}

// THIS CODE IS CONTRIBUTED BY YASH AGARWAL
JavaScript 
// JavaScript implementation of above approach
// a binary tree node
class Node{
    constructor(data){
        this.data = data;
        this.left = null;
        this.right = null;
    }
}

// utility functionto create a new node
function newNode(data){
    return new Node(data);
}

// function will return the deepest left leaf node
function deepestLeftLeaf(root){
    if(root == null) return null;
    let ans = null;
    let q = [];
    q.push(root);
    while(q.length > 0){
        let front_node = q.shift();
        if(front_node.left){
            q.push(front_node.left);
            ans = front_node.left;
        }
        if(front_node.right) q.push(front_node.right);
    }
    return ans;
}

// driver code to test above function
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 = deepestLeftLeaf(root);
if(result)
    console.log("The deepest left child is : " + result.data);
else
    console.log("There is no left leaf in the given tree.");

Output
The deepest left child is 9

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


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