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.


Input : x = 45,
   5 is Root of below tree
      /    \
    10      15
    / \    /  \
  20  25  30   35
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
    // pushing marker to the queue
    // Variable to store count of level
    int levelCount = 0;
    while (!q.empty()) {
        Node* temp = q.front();
        // if node is marker, push marker to queue
        // else, push left and right (if exists)
        if (temp == NULL && !q.empty()) {
            // Increment levelCount, while moving
            // to new level
        else if (temp != NULL) {
            // If node at current level is Key,
            // return levelCount
            if (temp->data == key)
                return levelCount;
            if (temp->left)
            if (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;




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