# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` ` `  `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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