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Print all the nodes except the leftmost node in every level of the given binary tree
• Difficulty Level : Easy
• Last Updated : 17 Jan, 2020

Given a binary tree, the task is to print all the nodes except the leftmost in every level of the tree. The root is considered at level 0, and left most node of any level is considered as a node at position 0.

Examples:

```Input:
1
/     \
2       3
/   \       \
4     5       6
/  \
7    8
/      \
9        10

Output:
3
5  6
8
10

Input:
1
/   \
2     3
\     \
4     5
Output:
3
5
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: To print nodes level by level, use level order traversal. The idea is based on Print level order traversal line by line. For that, traverse nodes level by level and mark leftmost flag true just before the processing of each level and mark it false just after processing of the first node at each level.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Structure of the tree node ` `struct` `Node { ` `    ``int` `data; ` `    ``Node *left, *right; ` `}; ` ` `  `// Utility method to create a node ` `struct` `Node* newNode(``int` `data) ` `{ ` `    ``struct` `Node* node = ``new` `Node; ` `    ``node->data = data; ` `    ``node->left = node->right = NULL; ` `    ``return` `(node); ` `} ` ` `  `// Function to print all the nodes ` `// except the leftmost in every level ` `// of the given binary tree ` `// with level order traversal ` `void` `excludeLeftmost(Node* root) ` `{ ` `    ``// Base Case ` `    ``if` `(root == NULL) ` `        ``return``; ` ` `  `    ``// Create an empty queue for level ` `    ``// order traversal ` `    ``queue q; ` ` `  `    ``// Enqueue root ` `    ``q.push(root); ` ` `  `    ``while` `(1) { ` ` `  `        ``// nodeCount (queue size) indicates ` `        ``// number of nodes at current level. ` `        ``int` `nodeCount = q.size(); ` `        ``if` `(nodeCount == 0) ` `            ``break``; ` ` `  `        ``// Initialize leftmost as true ` `        ``// just before the beginning ` `        ``// of each level ` `        ``bool` `leftmost = ``true``; ` ` `  `        ``// Dequeue all nodes of current level ` `        ``// and Enqueue all nodes of next level ` `        ``while` `(nodeCount > 0) { ` `            ``Node* node = q.front(); ` ` `  `            ``// Switch leftmost flag after processing ` `            ``// the leftmost node ` `            ``if` `(leftmost) ` `                ``leftmost = !leftmost; ` ` `  `            ``// Print all the nodes except leftmost ` `            ``else` `                ``cout << node->data << ``" "``; ` `            ``q.pop(); ` `            ``if` `(node->left != NULL) ` `                ``q.push(node->left); ` `            ``if` `(node->right != NULL) ` `                ``q.push(node->right); ` `            ``nodeCount--; ` `        ``} ` `        ``cout << ``"\n"``; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``struct` `Node* root = newNode(1); ` `    ``root->left = newNode(2); ` `    ``root->right = newNode(3); ` `    ``root->left->left = newNode(4); ` `    ``root->left->right = newNode(5); ` `    ``root->right->left = newNode(6); ` `    ``root->right->right = newNode(7); ` `    ``root->left->right->left = newNode(8); ` `    ``root->left->right->right = newNode(9); ` `    ``root->left->right->right->right = newNode(10); ` ` `  `    ``excludeLeftmost(root); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `import` `java.util.*; ` ` `  `class` `Sol ` `{ ` `     `  `// Structure of the tree node  ` `static` `class` `Node  ` `{  ` `    ``int` `data;  ` `    ``Node left, right;  ` `};  ` ` `  `// Utility method to create a node  ` `static` `Node newNode(``int` `data)  ` `{  ` `    ``Node node = ``new` `Node();  ` `    ``node.data = data;  ` `    ``node.left = node.right = ``null``;  ` `    ``return` `(node);  ` `}  ` ` `  `// Function to print all the nodes  ` `// except the leftmost in every level  ` `// of the given binary tree  ` `// with level order traversal  ` `static` `void` `excludeLeftmost(Node root)  ` `{  ` `    ``// Base Case  ` `    ``if` `(root == ``null``)  ` `        ``return``;  ` ` `  `    ``// Create an empty queue for level  ` `    ``// order traversal  ` `    ``Queue q = ``new` `LinkedList();  ` ` `  `    ``// Enqueue root  ` `    ``q.add(root);  ` ` `  `    ``while` `(``true``)  ` `    ``{  ` ` `  `        ``// nodeCount (queue size) indicates  ` `        ``// number of nodes at current level.  ` `        ``int` `nodeCount = q.size();  ` `        ``if` `(nodeCount == ``0``)  ` `            ``break``;  ` ` `  `        ``// Initialize leftmost as true  ` `        ``// just before the beginning  ` `        ``// of each level  ` `        ``boolean` `leftmost = ``true``;  ` ` `  `        ``// Dequeue all nodes of current level  ` `        ``// and Enqueue all nodes of next level  ` `        ``while` `(nodeCount > ``0``) ` `        ``{  ` `            ``Node node = q.peek();  ` ` `  `            ``// Switch leftmost flag after processing  ` `            ``// the leftmost node  ` `            ``if` `(leftmost)  ` `                ``leftmost = !leftmost;  ` ` `  `            ``// Print all the nodes except leftmost  ` `            ``else` `                ``System.out.print( node.data + ``" "``);  ` `            ``q.remove();  ` `            ``if` `(node.left != ``null``)  ` `                ``q.add(node.left);  ` `            ``if` `(node.right != ``null``)  ` `                ``q.add(node.right);  ` `            ``nodeCount--;  ` `        ``}  ` `        ``System.out.println();  ` `    ``}  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main(String args[]) ` `{  ` `    ``Node root = newNode(``1``);  ` `    ``root.left = newNode(``2``);  ` `    ``root.right = newNode(``3``);  ` `    ``root.left.left = newNode(``4``);  ` `    ``root.left.right = newNode(``5``);  ` `    ``root.right.left = newNode(``6``);  ` `    ``root.right.right = newNode(``7``);  ` `    ``root.left.right.left = newNode(``8``);  ` `    ``root.left.right.right = newNode(``9``);  ` `    ``root.left.right.right.right = newNode(``10``);  ` ` `  `    ``excludeLeftmost(root);  ` `} ` `}  ` ` `  `// This code is contributed by Arnab Kundu `

## Python3

 `# Python implementation of the approach ` `from` `collections ``import` `deque ` ` `  `# Structure of the tree node ` `class` `Node: ` `    ``def` `__init__(``self``): ` `        ``self``.data ``=` `0` `        ``self``.left ``=` `None` `        ``self``.right ``=` `None` ` `  `# Utility method to create a node ` `def` `newNode(data: ``int``) ``-``> Node: ` `    ``node ``=` `Node() ` `    ``node.data ``=` `data ` `    ``node.left ``=` `None` `    ``node.right ``=` `None` `    ``return` `node ` ` `  `# Function to print all the nodes ` `# except the leftmost in every level ` `# of the given binary tree ` `# with level order traversal ` `def` `excludeLeftMost(root: Node): ` ` `  `    ``# Base Case ` `    ``if` `root ``is` `None``: ` `        ``return` ` `  `    ``# Create an empty queue for level ` `    ``# order traversal ` `    ``q ``=` `deque() ` ` `  `    ``# Enqueue root ` `    ``q.append(root) ` ` `  `    ``while` `1``: ` ` `  `        ``# nodeCount (queue size) indicates ` `        ``# number of nodes at current level ` `        ``nodeCount ``=` `len``(q) ` `        ``if` `nodeCount ``=``=` `0``: ` `            ``break` ` `  `        ``# Initialize leftmost as true ` `        ``# just before the beginning ` `        ``# of each level ` `        ``leftmost ``=` `True` ` `  `        ``# Dequeue all nodes of current level ` `        ``# and Enqueue all nodes of next level ` `        ``while` `nodeCount > ``0``: ` `            ``node ``=` `q[``0``] ` ` `  `            ``# Switch leftmost flag after processing ` `            ``# the leftmost node ` `            ``if` `leftmost: ` `                ``leftmost ``=` `not` `leftmost ` ` `  `            ``# Print all the nodes except leftmost ` `            ``else``: ` `                ``print``(node.data, end``=``" "``) ` `            ``q.popleft() ` ` `  `            ``if` `node.left ``is` `not` `None``: ` `                ``q.append(node.left) ` `            ``if` `node.right ``is` `not` `None``: ` `                ``q.append(node.right) ` `            ``nodeCount ``-``=` `1` `        ``print``() ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` `    ``root ``=` `Node() ` `    ``root ``=` `newNode(``1``) ` `    ``root.left ``=` `newNode(``2``) ` `    ``root.right ``=` `newNode(``3``) ` `    ``root.left.left ``=` `newNode(``4``) ` `    ``root.left.right ``=` `newNode(``5``) ` `    ``root.right.left ``=` `newNode(``6``) ` `    ``root.right.right ``=` `newNode(``7``) ` `    ``root.left.right.left ``=` `newNode(``8``) ` `    ``root.left.right.right ``=` `newNode(``9``) ` `    ``root.left.right.right.right ``=` `newNode(``10``) ` `    ``excludeLeftMost(root) ` ` `  `# This code is contributed by ` `# sanjeev2552 `

## C#

 `// C# implementation of the above approach  ` `using` `System;  ` `using` `System.Collections.Generic;  ` ` `  `class` `GFG ` `{ ` `     `  `// Structure of the tree node  ` `public` `class` `Node  ` `{  ` `    ``public` `int` `data;  ` `    ``public` `Node left, right;  ` `};  ` ` `  `// Utility method to create a node  ` `static` `Node newNode(``int` `data)  ` `{  ` `    ``Node node = ``new` `Node();  ` `    ``node.data = data;  ` `    ``node.left = node.right = ``null``;  ` `    ``return` `(node);  ` `}  ` ` `  `// Function to print all the nodes  ` `// except the leftmost in every level  ` `// of the given binary tree  ` `// with level order traversal  ` `static` `void` `excludeLeftmost(Node root)  ` `{  ` `    ``// Base Case  ` `    ``if` `(root == ``null``)  ` `        ``return``;  ` ` `  `    ``// Create an empty queue for level  ` `    ``// order traversal  ` `    ``Queue q = ``new` `Queue();  ` ` `  `    ``// Enqueue root  ` `    ``q.Enqueue(root);  ` ` `  `    ``while` `(``true``)  ` `    ``{  ` ` `  `        ``// nodeCount (queue size) indicates  ` `        ``// number of nodes at current level.  ` `        ``int` `nodeCount = q.Count;  ` `        ``if` `(nodeCount == 0)  ` `            ``break``;  ` ` `  `        ``// Initialize leftmost as true  ` `        ``// just before the beginning  ` `        ``// of each level  ` `        ``Boolean leftmost = ``true``;  ` ` `  `        ``// Dequeue all nodes of current level  ` `        ``// and Enqueue all nodes of next level  ` `        ``while` `(nodeCount > 0) ` `        ``{  ` `            ``Node node = q.Peek();  ` ` `  `            ``// Switch leftmost flag after processing  ` `            ``// the leftmost node  ` `            ``if` `(leftmost)  ` `                ``leftmost = !leftmost;  ` ` `  `            ``// Print all the nodes except leftmost  ` `            ``else` `                ``Console.Write( node.data + ``" "``);  ` `            ``q.Dequeue();  ` `            ``if` `(node.left != ``null``)  ` `                ``q.Enqueue(node.left);  ` `            ``if` `(node.right != ``null``)  ` `                ``q.Enqueue(node.right);  ` `            ``nodeCount--;  ` `        ``}  ` `        ``Console.WriteLine();  ` `    ``}  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `Main(String []args) ` `{  ` `    ``Node root = newNode(1);  ` `    ``root.left = newNode(2);  ` `    ``root.right = newNode(3);  ` `    ``root.left.left = newNode(4);  ` `    ``root.left.right = newNode(5);  ` `    ``root.right.left = newNode(6);  ` `    ``root.right.right = newNode(7);  ` `    ``root.left.right.left = newNode(8);  ` `    ``root.left.right.right = newNode(9);  ` `    ``root.left.right.right.right = newNode(10);  ` ` `  `    ``excludeLeftmost(root);  ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992  `

Output:

```3
5 6 7
9
```

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