# 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++ 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; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

3

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