Find the Neighbours of a Node in a tree Last Updated : 23 Jul, 2025 Comments Improve Suggest changes Like Article Like Report Given a tree and a value of the node find out whether the node exists if exists, then find the neighbors of that node. Examples: Input: Consider the below tree as input and K = 4 Binary TreeOutput: 4 5 6 7 Input: Consider the below tree and K = 8 Binary TreeOutput: Node with Value 8 does not exist. Approach: To solve the problem follow the below steps: To check if given node exists or not: If the root pointer is nullptr, indicating an empty tree, return false because the node doesn't exist.Check if the value of the current root node is equal to the targetValue. If they are equal, return true.If the targetValue is not found at the current node, make two recursive calls:Call nodeExists on the left subtree (root -> left).Call nodeExists on the right subtree (root -> right).Find the neighoubours of a node using Level Order Traversal (Breadth First Search or BFS): Initializes two vectors ans to store nodes at each level during traversal and vec to store the neighbors of the target node.If the root is NULL, indicating an empty tree, the function returns an empty vectorFirst performs a level-order traversal of the tree using a queue. It processes nodes level by level.Now traverse the ans vector and checks if the current node's value matches the targetValue. If found, it stores the nodes at that level in vec.If neighbors are found, it returns vec containing the neighbors. If no neighbors are found, it returns an empty vector.Below is the implementation of above approach: C++ // C++ code for the above approach: #include <bits/stdc++.h> using namespace std; class TreeNode { public: int value; TreeNode* left; TreeNode* right; TreeNode(int val) : value(val) , left(nullptr) , right(nullptr) { } }; // Check if a node with a specific value // exists in a binary tree bool nodeExists(TreeNode* root, int targetValue) { // Node doesn't exist if (root == nullptr) { return false; } // Node exists if (root->value == targetValue) { return true; } // Recursive call return nodeExists(root->left, targetValue) || nodeExists(root->right, targetValue); } vector<int> findNodeNeighbors(TreeNode* root, int targetValue) { // Vector to store nodes at each level vector<vector<int> > ans; // Vector to store neighbours vector<int> vec; if (root == NULL) { return vec; } // To store the level of the node queue<TreeNode*> q; q.push(root); while (!q.empty()) { int size = q.size(); // Vector to store nodes at the // current level vector<int> ds; for (int i = 0; i < size; i++) { TreeNode* temp = q.front(); q.pop(); ds.push_back(temp->value); // If temp->left exists push if (temp->left != NULL) { q.push(temp->left); } // If temp->right exists push if (temp->right != NULL) { q.push(temp->right); } } // Storing the nodes at current level ans.push_back(ds); } // Finding the neighbours using for loop for (auto it : ans) { if (find(it.begin(), it.end(), targetValue) != it.end()) { vec = it; // Return neighbors if found return vec; } } // Return an empty vector if not found return vec; } // Driver code int main() { // Declaring tree TreeNode* root = new TreeNode(1); root->left = new TreeNode(2); root->right = new TreeNode(3); root->left->left = new TreeNode(4); root->left->right = new TreeNode(5); root->right->left = new TreeNode(6); root->right->right = new TreeNode(7); // For storing the answer vector<int> ans; // Value to find int targetValue = 4; if (nodeExists(root, targetValue)) { // Exist cout << "Node with value " << targetValue << " exists." << endl; ans = findNodeNeighbors(root, targetValue); } else { // Does not exist cout << "Node with value " << targetValue << " does not exist." << endl; } // Printing the neighbours cout << "Neighbours are:\n"; for (auto it : ans) { cout << it << " "; } // Deallocate memory delete root->left->left; delete root->left->right; delete root->left; delete root->right->left; delete root->right; delete root; return 0; } Java import java.util.ArrayList; import java.util.LinkedList; import java.util.List; import java.util.Queue; class TreeNode { public int value; public TreeNode left; public TreeNode right; public TreeNode(int val) { value = val; left = null; right = null; } } public class BinaryTree { // Check if a node with the target value exists in the binary tree public static boolean nodeExists(TreeNode root, int targetValue) { if (root == null) { return false; // If the root is null, the node does not exist. } if (root.value == targetValue) { return true; // If the current node's value matches the target, it exists. } // Recursively search in the left and right subtrees. return nodeExists(root.left, targetValue) || nodeExists(root.right, targetValue); } // Find neighbors of a node with the target value in the binary tree public static List<Integer> findNodeNeighbors(TreeNode root, int targetValue) { List<List<Integer>> ans = new ArrayList<>(); // List of lists to store levels of nodes List<Integer> vec = new ArrayList<>(); // List to store neighbors if (root == null) { return vec; // If the root is null, there are no neighbors. } Queue<TreeNode> q = new LinkedList<>(); q.add(root); while (!q.isEmpty()) { int size = q.size(); List<Integer> ds = new ArrayList<>(); // List to store nodes at the current level for (int i = 0; i < size; i++) { TreeNode temp = q.poll(); ds.add(temp.value); if (temp.left != null) { q.add(temp.left); } if (temp.right != null) { q.add(temp.right); } } ans.add(ds); // Store the nodes at the current level in the answer list. } for (List<Integer> it : ans) { if (it.contains(targetValue)) { vec = it; // If the target value is found in a level, set it as the neighbors. return vec; } } return vec; // Return the list of neighbors. } public static void main(String[] args) { TreeNode root = new TreeNode(1); root.left = new TreeNode(2); root.right = new TreeNode(3); root.left.left = new TreeNode(4); root.left.right = new TreeNode(5); root.right.left = new TreeNode(6); root.right.right = new TreeNode(7); List<Integer> ans = new ArrayList<>(); int targetValue = 4; if (nodeExists(root, targetValue)) { System.out.println("Node with value " + targetValue + " exists."); ans = findNodeNeighbors(root, targetValue); } else { System.out.println("Node with value " + targetValue + " does not exist."); } System.out.println("Neighbours are:"); for (int it : ans) { System.out.print(it + " "); } } } Python3 from collections import deque class TreeNode: def __init__(self, val): self.value = val self.left = None self.right = None # Check if a node with the target value exists in the binary tree def nodeExists(root, targetValue): if root is None: return False # If the root is None, the node does not exist. if root.value == targetValue: return True # If the current node's value matches the target, it exists. # Recursively search in the left and right subtrees. return nodeExists(root.left, targetValue) or nodeExists(root.right, targetValue) # Find neighbors of a node with the target value in the binary tree def findNodeNeighbors(root, targetValue): ans = [] # List of lists to store levels of nodes vec = [] # List to store neighbors if root is None: return vec # If the root is None, there are no neighbors. q = deque() q.append(root) while q: size = len(q) ds = [] # List to store nodes at the current level for i in range(size): temp = q.popleft() ds.append(temp.value) if temp.left: q.append(temp.left) if temp.right: q.append(temp.right) ans.append(ds) # Store the nodes at the current level in the answer list. for it in ans: if targetValue in it: vec = it # If the target value is found in a level, set it as the neighbors. return vec return vec # Return the list of neighbors. if __name__ == "__main__": root = TreeNode(1) root.left = TreeNode(2) root.right = TreeNode(3) root.left.left = TreeNode(4) root.left.right = TreeNode(5) root.right.left = TreeNode(6) root.right.right = TreeNode(7) ans = [] targetValue = 4 if nodeExists(root, targetValue): print("Node with value", targetValue, "exists.") ans = findNodeNeighbors(root, targetValue) else: print("Node with value", targetValue, "does not exist.") print("Neighbours are:") for it in ans: print(it, end=" ") #This code is contributed by chinmaya121221 C# using System; using System.Collections.Generic; using System.Linq; public class TreeNode { public int value; public TreeNode left; public TreeNode right; public TreeNode(int val) { value = val; left = null; right = null; } } public class BinaryTree { public static bool NodeExists(TreeNode root, int targetValue) { if (root == null) return false; if (root.value == targetValue) return true; return NodeExists(root.left, targetValue) || NodeExists(root.right, targetValue); } public static List<int> FindNodeNeighbors(TreeNode root, int targetValue) { var ans = new List<List<int>>(); var vec = new List<int>(); if (root == null) return vec; Queue<TreeNode> q = new Queue<TreeNode>(); q.Enqueue(root); while (q.Count > 0) { int size = q.Count; List<int> ds = new List<int>(); for (int i = 0; i < size; i++) { TreeNode temp = q.Dequeue(); ds.Add(temp.value); if (temp.left != null) q.Enqueue(temp.left); if (temp.right != null) q.Enqueue(temp.right); } ans.Add(ds); } foreach (var it in ans) { if (it.Contains(targetValue)) { vec = it; return vec; } } return vec; } public static void Main(string[] args) { TreeNode root = new TreeNode(1); root.left = new TreeNode(2); root.right = new TreeNode(3); root.left.left = new TreeNode(4); root.left.right = new TreeNode(5); root.right.left = new TreeNode(6); root.right.right = new TreeNode(7); List<int> ans = new List<int>(); int targetValue = 4; if (NodeExists(root, targetValue)) { Console.WriteLine($"Node with value {targetValue} exists."); ans = FindNodeNeighbors(root, targetValue); } else { Console.WriteLine($"Node with value {targetValue} does not exist."); } Console.WriteLine("Neighbours are:"); foreach (var item in ans) { Console.Write($"{item} "); } } } JavaScript class TreeNode { constructor(val) { this.value = val; this.left = null; this.right = null; } } function nodeExists(root, targetValue) { if (root === null) { return false; } if (root.value === targetValue) { return true; } return nodeExists(root.left, targetValue) || nodeExists(root.right, targetValue); } function findNodeNeighbors(root, targetValue) { const ans = []; if (root === null) { return []; } const queue = [root]; while (queue.length > 0) { const size = queue.length; const ds = []; for (let i = 0; i < size; i++) { const temp = queue.shift(); ds.push(temp.value); if (temp.left !== null) { queue.push(temp.left); } if (temp.right !== null) { queue.push(temp.right); } } ans.push(ds); } for (const it of ans) { if (it.includes(targetValue)) { return it; } } return []; } // Driver code const root = new TreeNode(1); root.left = new TreeNode(2); root.right = new TreeNode(3); root.left.left = new TreeNode(4); root.left.right = new TreeNode(5); root.right.left = new TreeNode(6); root.right.right = new TreeNode(7); const targetValue = 4; const ans = []; if (nodeExists(root, targetValue)) { console.log(`Node with value ${targetValue} exists.`); ans.push(...findNodeNeighbors(root, targetValue)); } else { console.log(`Node with value ${targetValue} does not exist.`); } console.log("Neighbours are:"); console.log(ans.join(" ")); // Deallocate memory (JavaScript automatically manages memory) OutputNode with value 4 exists. 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