Print leftmost and rightmost nodes of a Binary Tree

30 Jun, 2022
Given a Binary Tree, Print the corner nodes at each level. The node at the leftmost and the node at the rightmost. For example, output for following is 15, 10, 20, 8, 25. 
A Simple Solution is to do two traversals using the approaches discussed for printing left view and right view. 
Can we print all corner nodes using one traversal? The idea is to use Level Order Traversal. Every time we store the size of the queue in a variable n, which is the number of nodes at that level. For every level, we check whether the current node is the first (i.e node at index 0) and the node at the last index (i.e node at index n-1) If it is either of them, we print the value of that node.  
Implementation:
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// C/C++ program to print corner node at each level// of binary tree#include <bits/stdc++.h>using namespace std; /* A binary tree node has key, pointer to left   child and a pointer to right child */struct Node{    int key;    struct Node* left, *right;}; /* To create a newNode of tree and return pointer */struct Node* newNode(int key){    Node* temp = new Node;    temp->key = key;    temp->left = temp->right = NULL;    return (temp);} /* Function to print corner node at each level */void printCorner(Node *root){    //If the root is null then simply return    if(root == NULL)        return;    //Do level order traversal using a single queue    queue<Node*> q;    q.push(root);        while(!q.empty())    {        //n denotes the size of the current level in the queue        int n = q.size();                 for(int i =0;i<n;i++)        {            Node *temp = q.front();            q.pop();                         //If it is leftmost corner value or rightmost corner value then print it            if(i==0 || i==n-1)               cout<<temp->key<<" ";             //push the left and right children of the temp node            if(temp->left)                q.push(temp->left);            if(temp->right)                q.push(temp->right);        }    }}// Driver program to test above functionint main (){    Node *root =  newNode(15);    root->left = newNode(10);    root->right = newNode(20);    root->left->left = newNode(8);    root->left->right = newNode(12);    root->right->left = newNode(16);    root->right->right = newNode(25);    printCorner(root);    return 0;} // This code is contributed by Tapeshdua420.
// Java program to print corner node at each level in a binary tree import java.util.*; /* A binary tree node has key, pointer to left   child and a pointer to right child */class Node{    int key;    Node left, right;     public Node(int key)    {        this.key = key;        left = right = null;    }} class BinaryTree{    Node root;     /* Function to print corner node at each level */    void printCorner(Node root)    {        //  star node is for keeping track of levels        Queue<Node> q = new LinkedList<Node>();         // pushing root node and star node        q.add(root);        // Do level order traversal of Binary Tree        while (!q.isEmpty())        {            // n is the no of nodes in current Level            int n = q.size();            for(int i = 0 ; i < n ; i++){            // dequeue the front node from the queue            Node temp = q.peek();            q.poll();            //If it is leftmost corner value or rightmost corner value then print it            if(i==0 || i==n-1)                System.out.print(temp.key + "  ");            //push the left and right children of the temp node            if (temp.left != null)                q.add(temp.left);            if (temp.right != null)                q.add(temp.right);        }        }     }     // Driver program to test above functions    public static void main(String[] args)    {        BinaryTree tree = new BinaryTree();        tree.root = new Node(15);        tree.root.left = new Node(10);        tree.root.right = new Node(20);        tree.root.left.left = new Node(8);        tree.root.left.right = new Node(12);        tree.root.right.left = new Node(16);        tree.root.right.right = new Node(25);         tree.printCorner(tree.root);    }} // This code has been contributed by Utkarsh Choubey
# Python3 program to print corner# node at each level of binary treefrom collections import deque # A binary tree node has key, pointer to left# child and a pointer to right childclass Node:    def __init__(self, key):                 self.key = key        self.left = None        self.right = None # Function to print corner node at each leveldef printCorner(root: Node):     # If the root is null then simply return    if root == None:        return     # Do level order traversal    # using a single queue    q = deque()    q.append(root)     while q:         # n denotes the size of the current        # level in the queue        n = len(q)        for i in range(n):            temp = q[0]            q.popleft()             # If it is leftmost corner value or            # rightmost corner value then print it            if i == 0 or i == n - 1:                print(temp.key, end = " ")             # push the left and right children            # of the temp node            if temp.left:                q.append(temp.left)            if temp.right:                q.append(temp.right) # Driver Codeif __name__ == "__main__":         root = Node(15)    root.left = Node(10)    root.right = Node(20)    root.left.left = Node(8)    root.left.right = Node(12)    root.right.left = Node(16)    root.right.right = Node(25)         printCorner(root) # This code is contributed by sanjeev2552
// C# program to print corner node// at each level in a binary treeusing System;using System.Collections.Generic; /* A binary tree node has key, pointer to leftchild and a pointer to right child */public class Node{    public int key;    public Node left, right;     public Node(int key)    {        this.key = key;        left = right = null;    }} public class BinaryTree{    Node root;     /* Function to print corner node at each level */    void printCorner(Node root)    {        // star node is for keeping track of levels        Queue<Node> q = new Queue<Node>();         // pushing root node and star node        q.Enqueue(root);        // Do level order traversal of Binary Tree        while (q.Count != 0)        {            // n is the no of nodes in current Level            int n = q.Count;            for(int i = 0 ; i < n ; i++){                Node temp = q.Peek();                q.Dequeue();                //If it is leftmost corner value or rightmost corner value then print it                if(i==0||i==n-1)                    Console.Write(temp.key + " ");            //push the left and right children of the temp node                if (temp.left != null)                    q.Enqueue(temp.left);                if (temp.right != null)                    q.Enqueue(temp.right);             }        } }     // Driver code    public static void Main(String[] args)    {        BinaryTree tree = new BinaryTree();        tree.root = new Node(15);        tree.root.left = new Node(10);        tree.root.right = new Node(20);        tree.root.left.left = new Node(8);        tree.root.left.right = new Node(12);        tree.root.right.left = new Node(16);        tree.root.right.right = new Node(25);         tree.printCorner(tree.root);    }} // This code is contributed by Utkarsh Choubey
15 10 20 8 25 
Time Complexity: O(n) where n is number of nodes in Binary Tree.
This article is contributed by Utkarsh Choubey.
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