chore(CPlusPlus): add print all nodes at a level on trees (#696)

Co-authored-by: Ming Tsai <37890026+ming-tsai@users.noreply.github.com>
Co-authored-by: Rahul Rajeev Pillai <66192267+Ashborn-SM@users.noreply.github.com>

algorithms/CPlusPlus/README.md

@@ -140,6 +140,7 @@
 - [Finding the elements of a tree visible from top view](Trees/Top-View-Of-A-Tree.cpp)
 - [Binary Tree Implementation](Trees/binary-tree-implementation.cpp)
 - [Iterative Segment Tree](Trees/IterativeSegmentTree.cpp)
+- [Print all nodes at level k](Trees/print-all-nodes-at-a-level.cpp)
 - [Sum of right leaves](Trees/sum-of-right-leaves.cpp)
 
 ## Trie

algorithms/CPlusPlus/Trees/print-all-nodes-at-a-level.cpp

@@ -0,0 +1,83 @@
+/* Description - We have to print all nodes at a level 'k' of the tree.
+
+For example- If we are given the following tree, the nodes at level 1 are 1, nodes at level 2 are 2,3, nodes at level 3 are 7, 9 and nodes at level 4 are 21 and 11.
+(Note- The level starts from 1, i.e root is at level 1)
+
+            1          level 1
+          /   \
+         2     3       level 2
+      	      /  \    
+             7    9    level 3
+	    /      \
+	   21      11  level 4
+*/
+
+#include<bits/stdc++.h>
+using namespace std;
+
+typedef struct Node
+{
+	int data;
+	struct Node* left;
+	struct Node* right;
+	
+	Node(int val)
+	{
+		data= val;
+		left=right=NULL;
+	}
+}node;
+
+//function for level order traversal
+void levelorder(Node* root, int k)
+{
+	if(root==NULL)
+		return;
+		
+	queue<Node*> q;
+	
+	q.push(root);
+	int count=0;  //for calculating at which level we are.
+	
+	while(!q.empty())
+	{
+		count++;  //increment the value of count as level is incremented
+		int n=q.size();
+		
+		for(int i=0; i<n; i++)
+		{
+			Node* temp=q.front();
+			q.pop();
+			if(count==k) { //if level is equal to the required level, then print its nodes
+				cout<<temp->data<<" ";
+			}
+			if(temp->left!=NULL){
+				q.push(temp->left);
+			}
+			if(temp->right!=NULL){
+				q.push(temp->right);
+			}
+		}	
+	}
+}
+
+//main function
+int main()
+{
+	node*root=new node(1);
+	root->left=new node(2);
+	root->right=new node(3);
+	root->right->left=new node(7);
+	root->right->right=new node(9);
+	root->right->left->left=new node(21);
+	root->right->right->right=new node(11);
+
+	int k = 3;  // here we have taken k=3 (third level of tree)
+	
+	levelorder(root,k);   //calling level order function
+}
+
+// The output of the above program will be 7 9. Since k=3 and nodes at level 3 are 7 and 9.
+
+// Time Complexity: O(n) where n is the number of nodes in the binary tree 
+// Space Complexity: O(n) where n is the number of nodes in the binary tree