Deepest right leaf node in a binary tree | Iterative approach

Given a Binary Tree, find the deepest leaf node that is the right child of its parent. For example, consider the following tree. The deepest right leaf node is the node with the value 10.

Examples:

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

Output : 10

The idea is similar to Method 2 of level order traversal

Traverse the tree level by level and while pushing right child to queue, check if it is leaf node, if it’s leaf node, then update the result and since we are traversing the level by level, the last stored right leaf will be the deepest right leaf node. 

Implementation:

C++

// CPP program to find deepest right 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 right leaf node
// of binary tree
Node* getDeepestRightLeafNode(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();

        if (temp->left) {
            q.push(temp->left);
        }

        // Since we go level by level, the last
        // stored right leaf node is deepest one
        if (temp->right) {
            q.push(temp->right);
            if (!temp->right->left && !temp->right->right)
                result = 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->right = 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 = getDeepestRightLeafNode(root);
    if (result)
        cout << "Deepest Right Leaf Node :: "
             << result->data << endl;
    else
        cout << "No result, right leaf not found\n";
    return 0;
}

Java

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

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 right leaf node
// of binary tree
static Node getDeepestRightLeafNode(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.poll();


        if (temp.left != null)
        {
            q.add(temp.left);
        }

        // Since we go level by level, the last
        // stored right leaf node is deepest one
        if (temp.right != null)
        {
            q.add(temp.right);
            if (temp.right.left == null && temp.right.right == null)
                result = 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.right = 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 = getDeepestRightLeafNode(root);
    if (result != null)
        System.out.println("Deepest Right Leaf Node :: "
            + result.data);
    else
        System.out.println("No result, right leaf not found\n");
    }
}

/* This code is contributed by PrinciRaj1992 */

Python3

# Python3 program to find closest
# value in 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 level
# of given node
def getDeepestRightLeafNode(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)

        # Since we go level by level, the last
        # stored right leaf node is deepest one
        if (temp.right):
            q.append(temp.right)
            if (not temp.right.left and
                not temp.right.right):
                result = 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.right = 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 = getDeepestRightLeafNode(root)
    if result:
        print("Deepest Right Leaf Node ::",
                               result.data)
    else:
        print("No result, right leaf not found")

# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

C#

// C# program to find deepest right leaf
// node of binary tree
using System;
using System.Collections.Generic;

class GFG
{


// tree node
public 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 right leaf node
// of binary tree
static Node getDeepestRightLeafNode(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();


        if (temp.left != null)
        {
            q.Enqueue(temp.left);
        }

        // Since we go level by level, the last
        // stored right leaf node is deepest one
        if (temp.right != null)
        {
            q.Enqueue(temp.right);
            if (temp.right.left == null && temp.right.right == null)
                result = 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.right = 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 = getDeepestRightLeafNode(root);
    if (result != null)
        Console.WriteLine("Deepest Right Leaf Node :: "
            + result.data);
    else
        Console.WriteLine("No result, right leaf not found\n");
}
}

// This code is contributed by Princi Singh

Javascript

<script>

    // JavaScript program to find deepest right leaf
    // node of binary tree

    /* A binary tree node has data, pointer to
    left child and a pointer to right child */
    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 right leaf node
    // of binary tree
    function getDeepestRightLeafNode(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();


            if (temp.left != null)
            {
                q.push(temp.left);
            }

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

    // construct a tree
    let root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.left.right = 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 = getDeepestRightLeafNode(root);
    if (result != null)
        document.write("Deepest Right Leaf Node :: "
            + result.data);
    else
        document.write("No result, right leaf not found");

</script>

Output

Deepest Right Leaf Node :: 10

Time Complexity: O(n)

Auxiliary Space: O(n)  because using queue