Given a binary tree and a value K. The task is to print the k-th node in the diagonal traversal of the binary tree. If no such node exists then print -1.
Examples:
Input : 8 / \ 3 10 / / \ 1 6 14 / \ / 4 7 13 k = 5 Output : 6 Diagonal Traversal of the above tree is: 8 10 14 3 6 7 13 1 4 Input : 1 / \ 2 3 / \ 4 5 k = 7 Output : -1
Approach: The idea is to perform the diagonal traversal of the binary tree until K nodes are visited in the diagonal traversal. While traversing for each node visited decrement the value of variable K and return the current node when the value of K becomes zero. If the diagonal traversal does not contain at least K nodes, return -1.
Below is the implementation of the above approach:
// C++ program to print kth node // in the diagonal traversal of a binary tree #include <bits/stdc++.h> using namespace std;
// A binary tree node has data, pointer to left // child and a pointer to right child struct Node {
int data;
Node *left, *right;
}; // Helper function that allocates a new node Node* newNode( int data)
{ Node* node = new Node();
node->data = data;
node->left = node->right = NULL;
return (node);
} // Iterative function to print kth node // in diagonal traversal of binary tree int diagonalPrint(Node* root, int k)
{ // Base cases
if (root == NULL || k == 0)
return -1;
int ans = -1;
queue<Node*> q;
// Push root node
q.push(root);
// Push delimiter NULL
q.push(NULL);
while (!q.empty()) {
Node* temp = q.front();
q.pop();
if (temp == NULL) {
if (q.empty()) {
// If kth node exists then return
// the answer
if (k == 0)
return ans;
// If kth node doesnt exists
// then break from the while loop
else
break ;
}
q.push(NULL);
}
else {
while (temp) {
// If the required kth node
// has been found then return the answer
if (k == 0)
return ans;
k--;
// Update the value of variable ans
// each time
ans = temp->data;
if (temp->left)
q.push(temp->left);
temp = temp->right;
}
}
}
// If kth node doesnt exists then
// return -1
return -1;
} // Driver Code int main()
{ Node* root = newNode(8);
root->left = newNode(3);
root->right = newNode(10);
root->left->left = newNode(1);
root->left->right = newNode(6);
root->right->right = newNode(14);
root->right->right->left = newNode(13);
root->left->right->left = newNode(4);
root->left->right->right = newNode(7);
int k = 9;
cout << diagonalPrint(root, k);
return 0;
} |
// Java program to print kth node // in the diagonal traversal of a binary tree import java.util.*;
class GFG
{ // A binary tree node has data, pointer to left //child and a pointer to right child static class Node
{ int data;
Node left, right;
}; // Helper function that allocates a new node static Node newNode( int data)
{ Node node = new Node();
node.data = data;
node.left = node.right = null ;
return (node);
} // Iterative function to print kth node // in diagonal traversal of binary tree static int diagonalPrint(Node root, int k)
{ // Base cases
if (root == null || k == 0 )
return - 1 ;
int ans = - 1 ;
Queue<Node> q = new LinkedList<Node>();
// add root node
q.add(root);
// add delimiter null
q.add( null );
while (q.size() > 0 )
{
Node temp = q.peek();
q.remove();
if (temp == null )
{
if (q.size() == 0 )
{
// If kth node exists then return
// the answer
if (k == 0 )
return ans;
// If kth node doesnt exists
// then break from the while loop
else
break ;
}
q.add( null );
}
else {
while (temp != null )
{
// If the required kth node
// has been found then return the answer
if (k == 0 )
return ans;
k--;
// Update the value of variable ans
// each time
ans = temp.data;
if (temp.left!= null )
q.add(temp.left);
temp = temp.right;
}
}
}
// If kth node doesnt exists then
// return -1
return - 1 ;
} // Driver Code public static void main(String args[])
{ Node root = newNode( 8 );
root.left = newNode( 3 );
root.right = newNode( 10 );
root.left.left = newNode( 1 );
root.left.right = newNode( 6 );
root.right.right = newNode( 14 );
root.right.right.left = newNode( 13 );
root.left.right.left = newNode( 4 );
root.left.right.right = newNode( 7 );
int k = 9 ;
System.out.println( diagonalPrint(root, k));
} } // This code is contributed by Arnab Kundu |
# Python program to print kth node # in the diagonal traversal of a binary tree # Linked List node class Node:
def __init__( self , data):
self .data = data
self .left = None
self .right = None
# Helper function that allocates a new node def newNode(data) :
node = Node( 0 )
node.data = data
node.left = node.right = None
return (node)
# Iterative function to print kth node # in diagonal traversal of binary tree def diagonalPrint( root, k) :
# Base cases
if (root = = None or k = = 0 ) :
return - 1
ans = - 1
q = []
# append root node
q.append(root)
# append delimiter None
q.append( None )
while ( len (q) > 0 ):
temp = q[ 0 ]
q.pop( 0 )
if (temp = = None ):
if ( len (q) = = 0 ) :
# If kth node exists then return
# the answer
if (k = = 0 ) :
return ans
# If kth node doesnt exists
# then break from the while loop
else :
break
q.append( None )
else :
while (temp ! = None ):
# If the required kth node
# has been found then return the answer
if (k = = 0 ) :
return ans
k = k - 1
# Update the value of variable ans
# each time
ans = temp.data
if (temp.left ! = None ):
q.append(temp.left)
temp = temp.right
# If kth node doesnt exists then
# return -1
return - 1
# Driver Code root = newNode( 8 )
root.left = newNode( 3 )
root.right = newNode( 10 )
root.left.left = newNode( 1 )
root.left.right = newNode( 6 )
root.right.right = newNode( 14 )
root.right.right.left = newNode( 13 )
root.left.right.left = newNode( 4 )
root.left.right.right = newNode( 7 )
k = 9
print ( diagonalPrint(root, k))
# This code is contributed by Arnab Kundu |
// C# program to print kth node // in the diagonal traversal of a binary tree using System;
using System.Collections.Generic;
class GFG
{ // A binary tree node has data, pointer to left //child and a pointer to right child public class Node
{ public int data;
public Node left, right;
}; // Helper function that allocates a new node static Node newNode( int data)
{ Node node = new Node();
node.data = data;
node.left = node.right = null ;
return (node);
} // Iterative function to print kth node // in diagonal traversal of binary tree static int diagonalPrint(Node root, int k)
{ // Base cases
if (root == null || k == 0)
return -1;
int ans = -1;
Queue<Node> q = new Queue<Node>();
// Enqueue root node
q.Enqueue(root);
// Enqueue delimiter null
q.Enqueue( null );
while (q.Count > 0)
{
Node temp = q.Peek();
q.Dequeue();
if (temp == null )
{
if (q.Count == 0)
{
// If kth node exists then return
// the answer
if (k == 0)
return ans;
// If kth node doesnt exists
// then break from the while loop
else
break ;
}
q.Enqueue( null );
}
else
{
while (temp != null )
{
// If the required kth node
// has been found then return the answer
if (k == 0)
return ans;
k--;
// Update the value of variable ans
// each time
ans = temp.data;
if (temp.left!= null )
q.Enqueue(temp.left);
temp = temp.right;
}
}
}
// If kth node doesnt exists then
// return -1
return -1;
} // Driver Code public static void Main(String []args)
{ Node root = newNode(8);
root.left = newNode(3);
root.right = newNode(10);
root.left.left = newNode(1);
root.left.right = newNode(6);
root.right.right = newNode(14);
root.right.right.left = newNode(13);
root.left.right.left = newNode(4);
root.left.right.right = newNode(7);
int k = 9;
Console.WriteLine( diagonalPrint(root, k));
} } /* This code is contributed by PrinciRaj1992 */ |
<script> // JavaScript program to print kth node // in the diagonal traversal of a binary tree // A binary tree node has data, pointer to left //child and a pointer to right child class Node { constructor()
{
this .data = 0;
this .left = null ;
this .right = null ;
}
}; // Helper function that allocates a new node function newNode(data)
{ var node = new Node();
node.data = data;
node.left = node.right = null ;
return (node);
} // Iterative function to print kth node // in diagonal traversal of binary tree function diagonalPrint(root, k)
{ // Base cases
if (root == null || k == 0)
return -1;
var ans = -1;
var q = [];
// push root node
q.push(root);
// push delimiter null
q.push( null );
while (q.length > 0)
{
var temp = q[0];
q.shift();
if (temp == null )
{
if (q.length == 0)
{
// If kth node exists then return
// the answer
if (k == 0)
return ans;
// If kth node doesnt exists
// then break from the while loop
else
break ;
}
q.push( null );
}
else
{
while (temp != null )
{
// If the required kth node
// has been found then return the answer
if (k == 0)
return ans;
k--;
// Update the value of variable ans
// each time
ans = temp.data;
if (temp.left!= null )
q.push(temp.left);
temp = temp.right;
}
}
}
// If kth node doesnt exists then
// return -1
return -1;
} // Driver Code var root = newNode(8);
root.left = newNode(3); root.right = newNode(10); root.left.left = newNode(1); root.left.right = newNode(6); root.right.right = newNode(14); root.right.right.left = newNode(13); root.left.right.left = newNode(4); root.left.right.right = newNode(7); var k = 9;
document.write( diagonalPrint(root, k)); </script> |
4
Time Complexity: O(N), where N is the total number of nodes in the binary tree.
Auxiliary Space: O(N)