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Print all nodes at distance K from given node: Iterative Approach

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Given a binary tree, a target node and an integer K, the task is to find all the nodes that are at distance K from the given target node. 

Consider the above-given Tree, For the target node 12. 
Input: K = 1 
Output: 8 10 14

Input: K = 2 
Output: 4 20

Input: K = 3 
output: 22 

Approach: 
There are generally two cases for the nodes at a distance of K: 

  1. Node at a distance K is a child node of the target node.
  2. Node at a distance K is the ancestor of the target node.

The idea is to store the parent node of every node in a hash-map with the help of the Level-order traversal on the tree. Then, Simply Traverse the nodes from the Target node using Breadth-First Search on the left-child, right-child, and the parent node. At any instant when the distance of a node the from the target node is equal to K then print all the nodes of the queue.

Below is the implementation of the above approach: 

C++

// C++ implementation to print all
// the nodes from the given target
// node iterative approach
 
#include <bits/stdc++.h>
 
using namespace std;
 
// Structure of the Node
struct Node {
    int val;
    Node *left, *right;
};
 
// Map to store the parent
// node of every node of
// the given Binary Tree
unordered_map<Node*, Node*> um;
 
// Function to store all nodes
// parent in unordered_map
void storeParent(Node* root)
{
 
    // Make a queue to do level-order
    // Traversal and store parent
    // of each node in unordered map
    queue<Node*> q;
    q.push(root);
     
    // Loop to iterate until the
    // queue is not empty
    while (!q.empty()) {
        Node* p = q.front();
        q.pop();
         
        // Condition if the node is a
        /// root node that storing its
        // parent as NULL
        if (p == root) {
            um[p] = NULL;
        }
         
        // if left child exist of
        // popped out node then store
        // parent as value and node as key
        if (p->left) {
            um[p->left] = p;
            q.push(p->left);
        }
        if (p->right) {
            um[p->right] = p;
            q.push(p->right);
        }
    }
}
 
// Function to find the nodes
// at distance K from given node
void nodeAtDistK(Node* root,
           Node* target, int k)
{
    // Keep track of each node
    // which are visited so that
    // while doing BFS we will
    // not traverse it again
    unordered_set<Node*> s;
    int dist = 0;
    queue<Node*> q;
    q.push(target);
    s.insert(target);
     
    // Loop to iterate over the nodes
    // until the queue is not empty
    while (!q.empty()) {
 
        // if distance is equal to K
        // then we found a node in tree
        // which is distance K
        if (dist == k) {
            while (!q.empty()) {
                cout << q.front()->val << " ";
                q.pop();
            }
            // no need to traverse further since we found the answer
            break ;
        }
 
        // BFS on node's left,
        // right and parent node
        int size = q.size();
        for (int i = 0; i < size; i++) {
            Node* p = q.front();
            q.pop();
 
            // if the left of node is not
            // visited yet then push it in
            // queue and insert it in set as well
            if (p->left &&
                s.find(p->left) == s.end()) {
                q.push(p->left);
                s.insert(p->left);
            }
 
            // if the right of node is not visited
            // yet then push it in queue
            // and insert it in set as well
            if (p->right &&
                s.find(p->right) == s.end()) {
                q.push(p->right);
                s.insert(p->right);
            }
 
            // if the parent of node is not visited
            // yet then push it in queue and
            // insert it in set as well
            if (um[p] && s.find(um[p]) == s.end()) {
                q.push(um[p]);
                s.insert(um[p]);
            }
        }
        dist++;
    }
}
 
// Function to create a newnode
Node* newnode(int val)
{
    Node* temp = new Node;
    temp->val = val;
    temp->left = temp->right = NULL;
    return temp;
}
 
// Driver Code
int main()
{
    Node* root = newnode(20);
    root->left = newnode(8);
    root->right = newnode(22);
    root->right->left = newnode(5);
    root->right->right = newnode(8);
    root->left->left = newnode(4);
    root->left->left->left = newnode(25);
    root->left->right = newnode(12);
    root->left->right->left =
                   newnode(10);
    root->left->right->left->left =
                   newnode(15);
    root->left->right->left->right =
                   newnode(18);
    root->left->right->left->right->right =
                   newnode(23);
    root->left->right->right =
                   newnode(14);
    Node* target = root->left->right;
    storeParent(root);
    nodeAtDistK(root, target, 3);
    return 0;
}

                    

Java

// Java implementation to print all
// the nodes from the given target
// node iterative approach
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Queue;
 
class GFG{
 
// Structure of the Node
static class Node
{
    int val;
    Node left, right;
 
    public Node(int val)
    {
        this.val = val;
        this.left = this.right = null;
    }
};
 
// Map to store the parent
// node of every node of
// the given Binary Tree
static HashMap<Node, Node> um = new HashMap<>();
 
// Function to store all nodes
// parent in unordered_map
static void storeParent(Node root)
{
     
    // Make a queue to do level-order
    // Traversal and store parent
    // of each node in unordered map
    Queue<Node> q = new LinkedList<>();
    q.add(root);
 
    // Loop to iterate until the
    // queue is not empty
    while (!q.isEmpty())
    {
        Node p = q.poll();
 
        // Condition if the node is a
        /// root node that storing its
        // parent as NULL
        if (p == root)
        {
            um.put(p, null);
        }
 
        // if left child exist of
        // popped out node then store
        // parent as value and node as key
        if (p.left != null)
        {
            um.put(p.left, p);
            q.add(p.left);
        }
        if (p.right != null)
        {
            um.put(p.right, p);
            q.add(p.right);
        }
    }
}
 
// Function to find the nodes
// at distance K from given node
static void nodeAtDistK(Node root,
                        Node target, int k)
{
     
    // Keep track of each node
    // which are visited so that
    // while doing BFS we will
    // not traverse it again
    HashSet<Node> s = new HashSet<>();
    int dist = 0;
     
    Queue<Node> q = new LinkedList<>();
    q.add(target);
    s.add(target);
 
    // Loop to iterate over the nodes
    // until the queue is not empty
    while (!q.isEmpty())
    {
         
        // If distance is equal to K
        // then we found a node in tree
        // which is distance K
        if (dist == k)
        {
            while (!q.isEmpty())
            {
                System.out.print(q.peek().val + " ");
                q.poll();
            }
        }
 
        // BFS on node's left,
        // right and parent node
        int size = q.size();
        for(int i = 0; i < size; i++)
        {
            Node p = q.poll();
 
            // If the left of node is not
            // visited yet then add it in
            // queue and insert it in set as well
            if (p.left != null && !s.contains(p.left))
            {
                q.add(p.left);
                s.add(p.left);
            }
 
            // If the right of node is not visited
            // yet then add it in queue
            // and insert it in set as well
            if (p.right != null && !s.contains(p.right))
            {
                q.add(p.right);
                s.add(p.right);
            }
 
            // If the parent of node is not visited
            // yet then add it in queue and
            // insert it in set as well
            if (um.get(p) != null &&
                !s.contains(um.get(p)))
            {
                q.add(um.get(p));
                s.add(um.get(p));
            }
        }
        dist++;
    }
}
 
// Driver Code
public static void main(String[] args)
{
    Node root = new Node(20);
    root.left = new Node(8);
    root.right = new Node(22);
    root.right.left = new Node(5);
    root.right.right = new Node(8);
    root.left.left = new Node(4);
    root.left.left.left = new Node(25);
    root.left.right = new Node(12);
    root.left.right.left = new Node(10);
    root.left.right.left.left = new Node(15);
    root.left.right.left.right = new Node(18);
    root.left.right.left.right.right = new Node(23);
    root.left.right.right = new Node(14);
     
    Node target = root.left.right;
     
    storeParent(root);
     
    nodeAtDistK(root, target, 3);
}
}
 
// This code is contributed by sanjeev2552

                    

Python3

# Python3 implementation to print all
# the nodes from the given target
# node iterative approach
from collections import deque
 
# Structure of the Node
class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None
 
# Map to store the parent
# node of every node of
# the given Binary Tree
um = {}
 
# Function to store all nodes
# parent in unordered_map
def storeParent(root):
 
    # Make a queue to do level-order
    # Traversal and store parent
    # of each node in unordered map
    q = deque([root])
 
    # Loop to iterate until the
    # queue is not empty
    while q:
        p = q.popleft()
 
        # Condition if the node is a
        # root node that storing its
        # parent as NULL
        if p == root:
            um[p] = None
 
        # if left child exist of
        # popped out node then store
        # parent as value and node as key
        if p.left:
            um[p.left] = p
            q.append(p.left)
        if p.right:
            um[p.right] = p
            q.append(p.right)
 
# Function to find the nodes
# at distance K from given node
def nodeAtDistK(root, target, k):
 
    # Keep track of each node
    # which are visited so that
    # while doing BFS we will
    # not traverse it again
    s = set()
    dist = 0
 
    q = deque([target])
    s.add(target)
 
    # Loop to iterate over the nodes
    # until the queue is not empty
    while q:
 
        # If distance is equal to K
        # then we found a node in tree
        # which is distance K
        if dist == k:
            while q:
                print(q[0].val, end=' ')
                q.popleft()
 
        # BFS on node's left,
        # right and parent node
        size = len(q)
        for i in range(size):
            p = q.popleft()
 
            # If the left of node is not
            # visited yet then add it in
            # queue and insert it in set as well
            if p.left and p.left not in s:
                q.append(p.left)
                s.add(p.left)
 
            # If the right of node is not visited
            # yet then add it in queue
            # and insert it in set as well
            if p.right and p.right not in s:
                q.append(p.right)
                s.add(p.right)
 
            # If the parent of node is not visited
            # yet then add it in queue and
            # insert it in set as well
            if um[p] and um[p] not in s:
                q.append(um[p])
                s.add(um[p])
        dist += 1
 
# Driver Code
if __name__ == '__main__':
    root = Node(20)
    root.left = Node(8)
    root.right = Node(22)
    root.left.left = Node(5)
    root.left.left.left = Node(25);
    root.left.right =  Node(12);
    root.left.right.left =  Node(10);
    root.left.right.left.left =  Node(15);
    root.left.right.left.right =  Node(18);
    root.left.right.left.right.right = Node(23);
    root.left.right.right = Node(14);
     
    target = root.left.right
 
    # Store parent of each node
    storeParent(root)
 
    # Find all nodes at distance
    # K from given node
    nodeAtDistK(root, target, 3)
     
# This code is contributed by Potta Lokesh

                    

C#

// C# implementation to print all
// the nodes from the given target
// node iterative approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Structure of the Node
public class Node
{
    public int val;
    public Node left, right;
 
    public Node(int val)
    {
        this.val = val;
        this.left = this.right = null;
    }
};
 
// Map to store the parent
// node of every node of
// the given Binary Tree
static Dictionary<Node,
                  Node> um = new Dictionary<Node,
                                            Node>();
 
// Function to store all nodes
// parent in unordered_map
static void storeParent(Node root)
{
     
    // Make a queue to do level-order
    // Traversal and store parent
    // of each node in unordered map
    List<Node> q = new List<Node>();
    q.Add(root);
 
    // Loop to iterate until the
    // queue is not empty
    while (q.Count != 0)
    {
        Node p = q[0];
        q.RemoveAt(0);
         
        // Condition if the node is a
        /// root node that storing its
        // parent as NULL
        if (p == root)
        {
            um.Add(p, null);
        }
 
        // If left child exist of
        // popped out node then store
        // parent as value and node as key
        if (p.left != null)
        {
            um.Add(p.left, p);
            q.Add(p.left);
        }
        if (p.right != null)
        {
            um.Add(p.right, p);
            q.Add(p.right);
        }
    }
}
 
// Function to find the nodes
// at distance K from given node
static void nodeAtDistK(Node root,
                        Node target, int k)
{
     
    // Keep track of each node
    // which are visited so that
    // while doing BFS we will
    // not traverse it again
    HashSet<Node> s = new HashSet<Node>();
    int dist = 0;
     
    List<Node> q = new List<Node>();
    q.Add(target);
    s.Add(target);
 
    // Loop to iterate over the nodes
    // until the queue is not empty
    while (q.Count != 0)
    {
         
        // If distance is equal to K
        // then we found a node in tree
        // which is distance K
        if (dist == k)
        {
            while (q.Count != 0)
            {
                Console.Write(q[0].val + " ");
                q.RemoveAt(0);
            }
        }
 
        // BFS on node's left,
        // right and parent node
        int size = q.Count;
        for(int i = 0; i < size; i++)
        {
            Node p = q[0];
            q.RemoveAt(0);
 
            // If the left of node is not
            // visited yet then add it in
            // queue and insert it in set as well
            if (p.left != null && !s.Contains(p.left))
            {
                q.Add(p.left);
                s.Add(p.left);
            }
 
            // If the right of node is not visited
            // yet then add it in queue and insert
            // it in set as well
            if (p.right != null && !s.Contains(p.right))
            {
                q.Add(p.right);
                s.Add(p.right);
            }
 
            // If the parent of node is not visited
            // yet then add it in queue and
            // insert it in set as well
            if (um[p] != null &&
                !s.Contains(um[p]))
            {
                q.Add(um[p]);
                s.Add(um[p]);
            }
        }
        dist++;
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    Node root = new Node(20);
    root.left = new Node(8);
    root.right = new Node(22);
    root.right.left = new Node(5);
    root.right.right = new Node(8);
    root.left.left = new Node(4);
    root.left.left.left = new Node(25);
    root.left.right = new Node(12);
    root.left.right.left = new Node(10);
    root.left.right.left.left = new Node(15);
    root.left.right.left.right = new Node(18);
    root.left.right.left.right.right = new Node(23);
    root.left.right.right = new Node(14);
     
    Node target = root.left.right;
     
    storeParent(root);
     
    nodeAtDistK(root, target, 3);
}
}
 
// This code is contributed by Princi Singh

                    

Javascript

// JavaScript implementation to print all
// the nodes from the given target
// node iterative approach
// structure of the node
class Node{
    constructor(data){
        this.val = data;
        this.left = null;
        this.right = null;
    }
}
 
// Map to store the parent
// node of every node of
// the given Binary Tree
um = new Map();
 
// Function to store all nodes
// parent in unordered_map
function storeParent(root){
    // Make a queue to do level-order
    // Traversal and store parent
    // of each node in unordered map
    let q = [];
    q.push(root);
     
    // Loop to iterate until the
    // queue is not empty
    while (q.length > 0) {
        let p = q.shift();
          
        // Condition if the node is a
        /// root node that storing its
        // parent as NULL
        if (p == root) {
            um.set(p, null);
        }
          
        // if left child exist of
        // popped out node then store
        // parent as value and node as key
        if (p.left) {
            um.set(p.left,p);
            q.push(p.left);
        }
        if (p.right) {
            um.set(p.right, p);
            q.push(p.right);
        }
    }
}
 
// Function to find the nodes
// at distance K from given node
function nodeAtDistK(root, target, k){
    // Keep track of each node
    // which are visited so that
    // while doing BFS we will
    // not traverse it again
    let s = new Set();
    let dist = 0;
    let q = [];
     
    q.push(target);
    s.add(target);
      
    // Loop to iterate over the nodes
    // until the queue is not empty
    while (q.length > 0) {
        // if distance is equal to K
        // then we found a node in tree
        // which is distance K
        if (dist == k) {
            while (q.length > 0) {
                document.write(q.shift().val + " ");
            }
            // no need to traverse further since we found the answer
            break;
        }
  
        // BFS on node's left,
        // right and parent node
        let size = q.length;
        for (let i = 0; i < size; i++) {
            let p = q.shift();
             
            // if the left of node is not
            // visited yet then push it in
            // queue and insert it in set as well
            if (p.left != null && (s.has(p.left) == false)) {
                q.push(p.left);
                s.add(p.left);
            }
  
            // if the right of node is not visited
            // yet then push it in queue
            // and insert it in set as well
            if (p.right != null && (s.has(p.right) == false)) {
                q.push(p.right);
                s.add(p.right);
            }
  
            // if the parent of node is not visited
            // yet then push it in queue and
            // insert it in set as well
            if (um.get(p) != null && s.has(um.get(p)) == false) {
                q.push(um.get(p));
                s.add(um.get(p));
            }
        }
        dist++;
    }
}
 
// Function to create a newnode
function newnode(val){
    return new Node(val);
}
 
// driver code
let root = newnode(20)
root.left = newnode(8)
root.right = newnode(22)
root.right.left = newnode(5)
root.right.right = newnode(8)
root.left.left = newnode(4)
root.left.left.left = newnode(25)
root.left.right = newnode(12)
root.left.right.left = newnode(10)
root.left.right.left.left = newnode(15)
root.left.right.left.right = newnode(18)
root.left.right.left.right.right = newnode(23)
root.left.right.right = newnode(14)
let target = root.left.right;
 
storeParent(root);
nodeAtDistK(root,target, 3);
 
// THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002)

                    

Output
23 25 22 

Complexity Analysis:

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


Last Updated : 10 Feb, 2023
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