Find Level wise positions of given node in a given Binary Tree

Last Updated : 01 Apr, 2022

Given a binary tree and an integer X, the task is to find out all the occurrences of X in the given tree, and print its level and its position from left to right on that level. If X is not found, print -1.

Examples:

Input: X=35
10
/ \
20 30
/ \ / \
40 60 35 50
Output: [(3, 3)]
Explanation: Integer X=35 is present in level 3 and its position is 3 from left.

Input: X= 2
1
/ \
2 3
/ \ / \
4 6 8 2
Output: [(2, 1), (3, 4)]
Explanation: Integer X=2 is present in level 2 and its position is 1 from left also it is present in level 3 and position from left is 4.

Approach: This problem can be solved using level order traversal of binary tree using queue.

Perform level order traversal of given Binary Tree using Queue.
Keep track of level count and the count of nodes from left to right during traversal
Whenever X is encountered during traversal, print or store the level count and position of current occurrence of X.

Below is the implementation of the above approach:

C++

// C++ program to print level and // position of Node X in binary tree  #include <bits/stdc++.h> using namespace std;  // A Binary Tree Node struct Node {     int data;     struct Node *left, *right; };  // Function to find and print // level and position of node X void printLevelandPosition(Node* root, int X) {     // Base Case     if (root == NULL)         return;      // Create an empty queue     queue<Node*> q;      // Enqueue Root and initialize height     q.push(root);      // Create and initialize current     // level and position by 1     int currLevel = 1, position = 1;      while (q.empty() == false) {          int size = q.size();          while (size--) {              Node* node = q.front();              // print if node data equal to X             if (node->data == X)                 cout << "(" << currLevel                      << " " << position                      << "), ";             q.pop();              // increment the position             position++;              // Enqueue left child             if (node->left != NULL)                 q.push(node->left);              // Enqueue right child             if (node->right != NULL)                 q.push(node->right);         }         // increment the level         currLevel++;         position = 1;     } }  // Utility function to create a new tree node Node* newNode(int data) {     Node* temp = new Node;     temp->data = data;     temp->left = temp->right = NULL;     return temp; }  // Driver program to test above functions int main() {     // Let us create binary tree     Node* root = newNode(1);     root->left = newNode(2);     root->right = newNode(3);     root->left->left = newNode(4);     root->left->right = newNode(2);      int X = 2;      printLevelandPosition(root, X);     return 0; }

Java

// JAVA program to print level and // position of Node X in binary tree import java.util.*;  // A Binary Tree Node class Node {   int data;   Node left;   Node right; } class GFG {    // Function to find and print   // level and position of node X   public static void printLevelandPosition(Node root,                                            int X)   {     // Base Case     if (root == null)       return;      // Create an empty queue     Queue<Node> q = new LinkedList<>();      // Enqueue Root and initialize height     q.add(root);      // Create and initialize current     // level and position by 1     int currLevel = 1, position = 1;      while (q.size() != 0) {        int size = q.size();        while ((size--) != 0) {          Node node = q.peek();          // print if node data equal to X         if (node.data == X)           System.out.print("(" + currLevel + " "                            + position + "), ");         q.remove();          // increment the position         position++;          // Enqueue left child         if (node.left != null)           q.add(node.left);          // Enqueue right child         if (node.right != null)           q.add(node.right);       }       // increment the level       currLevel++;       position = 1;     }   }    // Utility function to create a new tree node   public static Node newNode(int data)   {     Node temp = new Node();     temp.data = data;     temp.left = temp.right = null;     return temp;   }    // Driver program to test above functions   public static void main(String[] args)   {     // Let us create binary tree     Node root = newNode(1);     root.left = newNode(2);     root.right = newNode(3);     root.left.left = newNode(4);     root.left.right = newNode(2);      int X = 2;      printLevelandPosition(root, X);   } }  // This code is contributed by Taranpreet

Python3

# Python code for the above approach class Node:     def __init__(self,d):         self.data = d         self.left = None         self.right = None      # Function to find and print # level and position of node X def printLevelandPosition(root, X):       # Base Case     if (root == None):         return      # Create an empty queue     q = []      # Enqueue Root and initialize height     q.append(root)      # Create and initialize current     # level and position by 1     currLevel,position = 1,1      while (len(q) != 0):          size = len(q)          while (size != 0):              node = q[0]             q = q[1:]              # print if node data equal to X             if (node.data == X):                 print (f"({currLevel} {position}),",end =" ")                       # increment the position             position += 1              # Enqueue left child             if (node.left != None):                 q.append(node.left)              # Enqueue right child             if (node.right != None):                 q.append(node.right)                          size -= 1          # increment the level         currLevel += 1         position = 1  # Driver program to test above functions  # Let us create binary tree root = Node(1) root.left = Node(2) root.right = Node(3) root.left.left = Node(4) root.left.right = Node(2)  X = 2  printLevelandPosition(root, X)  # This code is contributed by shinjanpatra

// C# program to print level and // position of Node X in binary tree using System; using System.Collections.Generic;   // A Binary Tree Node public class Node {   public int data;   public Node left;   public Node right; }  public class GFG {    // Function to find and print   // level and position of node X   public static void printLevelandPosition(Node root, int X) {     // Base Case     if (root == null)       return;      // Create an empty queue     Queue<Node> q = new Queue<Node>();      // Enqueue Root and initialize height     q.Enqueue(root);      // Create and initialize current     // level and position by 1     int currLevel = 1, position = 1;      while (q.Count != 0) {        int size = q.Count;        while ((size--) != 0) {          Node node = q.Peek();          // print if node data equal to X         if (node.data == X)           Console.Write("(" + currLevel + " " + position + "), ");         q.Dequeue();          // increment the position         position++;          // Enqueue left child         if (node.left != null)           q.Enqueue(node.left);          // Enqueue right child         if (node.right != null)           q.Enqueue(node.right);       }       // increment the level       currLevel++;       position = 1;     }   }    // Utility function to create a new tree node   public static Node newNode(int data) {     Node temp = new Node();     temp.data = data;     temp.left = temp.right = null;     return temp;   }    // Driver program to test above functions   public static void Main(String[] args) {     // Let us create binary tree     Node root = newNode(1);     root.left = newNode(2);     root.right = newNode(3);     root.left.left = newNode(4);     root.left.right = newNode(2);      int X = 2;      printLevelandPosition(root, X);   } }  // This code is contributed by Rajput-Ji

JavaScript

<script>         // JavaScript code for the above approach         class Node {             constructor(d) {                 this.data = d;                 this.left = null;                 this.right = null;             }         }          // Function to find and print         // level and position of node X         function printLevelandPosition(root, X)          {                      // Base Case             if (root == null)                 return;              // Create an empty queue             let q = [];              // Enqueue Root and initialize height             q.push(root);              // Create and initialize current             // level and position by 1             let currLevel = 1, position = 1;              while (q.length != 0) {                  let size = q.length;                  while (size-- != 0) {                      let node = q.shift();                      // print if node data equal to X                     if (node.data == X)                         document.write("(" + currLevel                             + " " + position                             + "), ");                       // increment the position                     position++;                      // Enqueue left child                     if (node.left != null)                         q.push(node.left);                      // Enqueue right child                     if (node.right != null)                         q.push(node.right);                 }                 // increment the level                 currLevel++;                 position = 1;             }         }          // Driver program to test above functions          // Let us create binary tree         let root = new Node(1);         root.left = new Node(2);         root.right = new Node(3);         root.left.left = new Node(4);         root.left.right = new Node(2);          let X = 2;          printLevelandPosition(root, X);         // This code is contributed by Potta Lokesh     </script>

Output

(2 1), (3 2),

Time Complexity: O(n)
Auxiliary Space : O(n)

Find if given vertical level of binary tree is sorted or not

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Find Level wise positions of given node in a given Binary Tree

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