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Print leftmost and rightmost nodes of a Binary Tree
  • Difficulty Level : Medium
  • Last Updated : 11 Aug, 2020

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 last index (i.e node at index n-1) , If it is either of them , we print the value of that node .

C/C++

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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 function
int 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 Utkarsh Choubey

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Java

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// 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

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Python3

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# Python3 program to print corner 
# node at each level of binary tree
from collections import deque
  
# A binary tree node has key, pointer to left
# child and a pointer to right child
class Node:
    def __init__(self, key):
          
        self.key = key
        self.left = None
        self.right = None
  
# Function to print corner node at each level
def 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 Code
if __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

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C#

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// C# program to print corner node 
// at each level in a binary tree
using System;
using System.Collections.Generic;
  
/* A binary tree node has key, pointer to left
child 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

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Output :



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. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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