Iterative program to find distance of a node from root

• Difficulty Level : Basic
• Last Updated : 29 Jun, 2021

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 using namespace std; // A Binary Tree Nodestruct Node {    int data;    Node *left, *right;}; // A utility function to create a new Binary// Tree NodeNode* 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 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 Codeint 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 Nodestatic class Node{    int data;    Node left, right;}; // A utility function to create a new Binary// Tree Nodestatic 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 q = new LinkedList();     // 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 Codepublic 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 nodeclass 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 calculateddef 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        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 Codeif __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 Nodeclass Node{    public int data;    public Node left, right;}; // A utility function to create a new Binary// Tree Nodestatic 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 q = new Queue();     // 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 Codepublic 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


Output:
3

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