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

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++ implementation to print all // the nodes from the given target // node iterative approach   #include   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 um;   // Functiom 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 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 give 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 s;     int dist = 0;     queue 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();             }         }           // 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; }

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