Iterative program to find distance of a node from root

Given the root of a binary tree and a key x in it, find the distance of the given key from the root node. Dis­tance means num­ber of edges between two nodes.

Examples: 

Input : x = 45,
   5 is Root of below tree
        5
      /    \
    10      15
    / \    /  \
  20  25  30   35
       \
       45
Output : Distance = 3             
There are three edges on path
from root to 45.

For more understanding of question,
in above tree distance of 35 is two
and distance of 10 is 1.

Related Problem: Recursive program to find distance of node from root.

Iterative Approach : 

• Use level order traversal to traverse the tree iteratively using a queue.
• Keep a variable levelCount to maintain the track of current level.
• To do this, every time on moving to the next level, while pushing a NULL node to the queue also increment the value of the variable levelCount so that it stores the current level number.
• While traversing the tree, check if any node at the current level matches with the given key.
• If yes, then return levelCount.

Below is the implementation of above approach:  

C++

// C++ program to find distance of a given
// node from root.
#include <bits/stdc++.h>
using namespace std;

// A Binary Tree Node
struct Node {
    int data;
    Node *left, *right;
};

// A utility function to create a new Binary
// Tree Node
Node* newNode(int item)
{
    Node* temp = new Node;
    temp->data = item;
    temp->left = temp->right = NULL;
    return temp;
}

/* Function to find distance of a node from root
*  root : root of the Tree
*  key : data whose distance to be calculated
*/
int findDistance(Node* root, int key)
{

    // base case
    if (root == NULL) {
        return -1;
    }

    // If the key is present at root,
    // distance is zero
    if (root->data == key)
        return 0;

    // Iterating through tree using BFS
    queue<Node*> q;

    // pushing root to the queue
    q.push(root);

    // pushing marker to the queue
    q.push(NULL);

    // Variable to store count of level
    int levelCount = 0;

    while (!q.empty()) {

        Node* temp = q.front();
        q.pop();

        // if node is marker, push marker to queue
        // else, push left and right (if exists)
        if (temp == NULL && !q.empty()) {
            q.push(NULL);

            // Increment levelCount, while moving
            // to new level
            levelCount++;
        }
        else if (temp != NULL) {

            // If node at current level is Key,
            // return levelCount
            if (temp->data == key)
                return levelCount;

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

            if (temp->right)
                q.push(temp->right);
        }
    }

    // If key is not found
    return -1;
}

// Driver Code
int main()
{
    Node* root = newNode(5);
    root->left = newNode(10);
    root->right = newNode(15);
    root->left->left = newNode(20);
    root->left->right = newNode(25);
    root->left->right->right = newNode(45);
    root->right->left = newNode(30);
    root->right->right = newNode(35);

    cout << findDistance(root, 45);

    return 0;
}

Java

// Java program to find distance of a given
// node from root.
import java.util.*;

class GFG
{

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

// A utility function to create a new Binary
// Tree Node
static Node newNode(int item)
{
    Node temp = new Node();
    temp.data = item;
    temp.left = temp.right = null;
    return temp;
}

/* Function to find distance of a node from root
* root : root of the Tree
* key : data whose distance to be calculated
*/
static int findDistance(Node root, int key)
{

    // base case
    if (root == null)
    {
        return -1;
    }

    // If the key is present at root,
    // distance is zero
    if (root.data == key)
        return 0;

    // Iterating through tree using BFS
    Queue<Node> q = new LinkedList<Node>();

    // adding root to the queue
    q.add(root);

    // adding marker to the queue
    q.add(null);

    // Variable to store count of level
    int levelCount = 0;

    while (!q.isEmpty())
    {
        Node temp = q.peek();
        q.remove();

        // if node is marker, push marker to queue
        // else, push left and right (if exists)
        if (temp == null && !q.isEmpty())
        {
            q.add(null);

            // Increment levelCount, while moving
            // to new level
            levelCount++;
        }

        else if (temp != null)
        {

            // If node at current level is Key,
            // return levelCount
            if (temp.data == key)
                return levelCount;

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

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

    // If key is not found
    return -1;
}

// Driver Code
public static void main(String[] args)
{
    Node root = newNode(5);
    root.left = newNode(10);
    root.right = newNode(15);
    root.left.left = newNode(20);
    root.left.right = newNode(25);
    root.left.right.right = newNode(45);
    root.right.left = newNode(30);
    root.right.right = newNode(35);

    System.out.println(findDistance(root, 45));
}
}

// This code is contributed by Rajput-Ji

Python3

# Python program to find distance of a given
# node from root.
from collections import deque

# A tree binary node
class Node:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None

# Function to find distance of a node from root
# root : root of the Tree
# key : data whose distance to be calculated
def findDistance(root: Node, key: int) -> int:

    # base case
    if root is None:
        return -1

    # If the key is present at root,
    # distance is zero
    if root.data == key:
        return 0

    # Iterating through tree using BFS
    q = deque()

    # pushing root to the queue
    q.append(root)

    # pushing marker to the queue
    q.append(None)

    # Variable to store count of level
    levelCount = 0

    while q:
        temp = q[0]
        q.popleft()

        # if node is marker, push marker to queue
        # else, push left and right (if exists)
        if temp is None and q:
            q.append(None)

            # Increment levelCount, while moving
            # to new level
            levelCount += 1
        elif temp:

            # If node at current level is Key,
            # return levelCount
            if temp.data == key:
                return levelCount

            if temp.left:
                q.append(temp.left)

            if temp.right:
                q.append(temp.right)

    # If key is not found
    return -1

# Driver Code
if __name__ == "__main__":

    root = Node(5)
    root.left = Node(10)
    root.right = Node(15)
    root.left.left = Node(20)
    root.left.right = Node(25)
    root.left.right.right = Node(45)
    root.right.left = Node(30)
    root.right.right = Node(35)

    print(findDistance(root, 45))

# This code is contributed by
# sanjeev2552

C#

// C# program to find distance of a given
// node from root.
using System;
using System.Collections.Generic;

class GFG
{

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

// A utility function to create a new Binary
// Tree Node
static Node newNode(int item)
{
    Node temp = new Node();
    temp.data = item;
    temp.left = temp.right = null;
    return temp;
}

/* Function to find distance of a node from root
* root : root of the Tree
* key : data whose distance to be calculated*/
static int findDistance(Node root, int key)
{

    // base case
    if (root == null)
    {
        return -1;
    }

    // If the key is present at root,
    // distance is zero
    if (root.data == key)
        return 0;

    // Iterating through tree using BFS
    Queue<Node> q = new Queue<Node>();

    // adding root to the queue
    q.Enqueue(root);

    // adding marker to the queue
    q.Enqueue(null);

    // Variable to store count of level
    int levelCount = 0;

    while (q.Count!=0)
    {
        Node temp = q.Peek();
        q.Dequeue();

        // if node is marker, push marker to queue
        // else, push left and right (if exists)
        if (temp == null && q.Count!=0)
        {
            q.Enqueue(null);

            // Increment levelCount, while moving
            // to new level
            levelCount++;
        }

        else if (temp != null)
        {

            // If node at current level is Key,
            // return levelCount
            if (temp.data == key)
                return levelCount;

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

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

    // If key is not found
    return -1;
}

// Driver Code
public static void Main(String[] args)
{
    Node root = newNode(5);
    root.left = newNode(10);
    root.right = newNode(15);
    root.left.left = newNode(20);
    root.left.right = newNode(25);
    root.left.right.right = newNode(45);
    root.right.left = newNode(30);
    root.right.right = newNode(35);

    Console.WriteLine(findDistance(root, 45));
}
}

// This code is contributed by Princi Singh

Javascript

<script>

// Javascript program to find distance
// of a given node from root.

// A Binary Tree Node
class Node
{
    constructor(item)
    {
        this.left = null;
        this.right = null;
        this.data = item;
    }
}

// A utility function to create a new Binary
// Tree Node
function newNode(item)
{
    let temp = new Node(item);
    return temp;
}

/* Function to find distance of a node from root
* root : root of the Tree
* key : data whose distance to be calculated
*/
function findDistance(root, key)
{

    // Base case
    if (root == null)
    {
        return -1;
    }

    // If the key is present at root,
    // distance is zero
    if (root.data == key)
        return 0;

    // Iterating through tree using BFS
    let q = [];

    // Adding root to the queue
    q.push(root);

    // Adding marker to the queue
    q.push(null);

    // Variable to store count of level
    let levelCount = 0;

    while (q.length > 0)
    {
        let temp = q[0];
        q.shift();

        // If node is marker, push marker to queue
        // else, push left and right (if exists)
        if (temp == null && q.length > 0)
        {
            q.push(null);

            // Increment levelCount, while moving
            // to new level
            levelCount++;
        }

        else if (temp != null)
        {

            // If node at current level is Key,
            // return levelCount
            if (temp.data == key)
                return levelCount;

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

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

    // If key is not found
    return -1;
}

// Driver code
let root = newNode(5);
root.left = newNode(10);
root.right = newNode(15);
root.left.left = newNode(20);
root.left.right = newNode(25);
root.left.right.right = newNode(45);
root.right.left = newNode(30);
root.right.right = newNode(35);

document.write(findDistance(root, 45));

// This code is contributed by suresh07

</script>

Output

3

Time Complexity: O(n) where n = Number of nodes
Space Complexity: O(n)