Check if a pair of leaf node sum exists in given binary tree
Here given code implementation process.
import java.util.HashMap;
/*
Java Program
Check if a pair of leaf node sum exists in given binary tree
*/
// Binary Tree node
class TreeNode
{
public int data;
public TreeNode left;
public TreeNode right;
public TreeNode(int data)
{
// Set node value
this.data = data;
this.left = null;
this.right = null;
}
}
public class BinaryTree
{
public TreeNode root;
public BinaryTree()
{
// Set initial tree root to null
this.root = null;
}
// Find all leaf nodes using recursion
public void findLeafNodes(TreeNode node,
HashMap <Integer,Integer> record)
{
if (node == null)
{
return;
}
if (node.left == null && node.right == null)
{
if (record.containsKey(node.data))
{
// Increment the element frequency
record.put(node.data,record.get(node.data)+1);
}
else
{
// Add new leaf node
record.put(node.data,1);
}
}
else
{
findLeafNodes(node.left, record);
findLeafNodes(node.right, record);
}
}
// Handles the request of check pair of leaf node sum
public void leafPairSum(int k)
{
if (this.root == null)
{
// Empty Tree
return;
}
else
{
// This is use to collect leaf nodes
HashMap < Integer, Integer > record =
new HashMap < Integer, Integer > ();
// Find all leaf nodes
findLeafNodes(this.root, record);
if (record.size() > 1)
{
System.out.print("\n Given sum k : " + k);
for (int key: record.keySet())
{
// Check that pair exist in tree
if (record.containsKey(k - key) &&
( (key != k-key) || record.get(k-key) > 1 ))
{
// Remaining amount exists in record and they are unique
// Display a pair of given sum
System.out.print("\n (" + key + " + " + (k - key) + ")");
return;
}
}
System.out.print("\n No leaf pair of given sum " + k);
}
else
{
System.out.print("\nSingle leaf node exists\n");
}
}
}
public static void main(String[] args)
{
// Create new binary tree
BinaryTree tree = new BinaryTree();
/*
4
/ \
9 7
/ \ \
-2 5 12
/ \ / \
6 8 5 7
/ / \
9 3 15
/ \ \
1 10 2
-----------------
Constructing binary tree
*/
tree.root = new TreeNode(4);
tree.root.left = new TreeNode(9);
tree.root.left.right = new TreeNode(5);
tree.root.left.right.left = new TreeNode(6);
tree.root.left.right.left.left = new TreeNode(9);
tree.root.left.right.right = new TreeNode(8);
tree.root.left.right.right.left = new TreeNode(3);
tree.root.left.right.right.left.left = new TreeNode(1);
tree.root.left.right.right.left.right = new TreeNode(10);
tree.root.left.left = new TreeNode(-2);
tree.root.right = new TreeNode(7);
tree.root.right.right = new TreeNode(12);
tree.root.right.right.right = new TreeNode(7);
tree.root.right.right.left = new TreeNode(5);
tree.root.right.right.left.right = new TreeNode(15);
tree.root.right.right.left.right.right = new TreeNode(2);
// Test cases
tree.leafPairSum(8);
tree.leafPairSum(2);
tree.leafPairSum(4);
tree.leafPairSum(-1);
tree.leafPairSum(9);
}
}input
Given sum k : 8
(-2 + 10)
Given sum k : 2
No leaf pair of given sum 2
Given sum k : 4
No leaf pair of given sum 4
Given sum k : -1
(-2 + 1)
Given sum k : 9
(2 + 7)// Include header file
#include <iostream>
#include <unordered_map>
using namespace std;
/*
C++ Program
Check if a pair of leaf node sum exists in given binary tree
*/
// Binary Tree node
class TreeNode
{
public: int data;
TreeNode *left;
TreeNode *right;
TreeNode(int data)
{
// Set node value
this->data = data;
this->left = NULL;
this->right = NULL;
}
};
class BinaryTree
{
public: TreeNode *root;
BinaryTree()
{
this->root = NULL;
}
// Find all leaf nodes using recursion
void findLeafNodes(TreeNode *node, unordered_map < int, int > &record)
{
if (node == NULL)
{
return;
}
if (node->left == NULL && node->right == NULL)
{
if (record.find(node->data) != record.end())
{
// Increment the element frequency
record[node->data] = record[node->data] + 1;
}
else
{
// Add new leaf node
record[node->data] = 1;
}
}
else
{
this->findLeafNodes(node->left, record);
this->findLeafNodes(node->right, record);
}
}
// Handles the request of check pair of leaf node sum
void leafPairSum(int k)
{
if (this->root == NULL)
{
// Empty Tree
return;
}
else
{
// This is use to collect leaf nodes
unordered_map < int, int > record;
// Find all leaf nodes
this->findLeafNodes(this->root, record);
if ( record.size() > 1)
{
cout << "\n Given sum k : " << k;
for (auto &info: record)
{
// Check that pair exist in tree
if (record.find(k - info.first) != record.end()
&& ((info.first != k - info.first)
|| record[k - info.first] > 1))
{
// Remaining amount exists in record and they are unique
// Display a pair of given sum
cout << "\n (" << info.first << " + "
<< (k - info.first) << ")";
return;
}
}
cout << "\n No leaf pair of given sum " << k;
}
else
{
cout << "\nSingle leaf node exists\n";
}
}
}
};
int main()
{
// Create new binary tree
BinaryTree *tree = new BinaryTree();
/*
4
/ \
9 7
/ \ \
-2 5 12
/ \ / \
6 8 5 7
/ / \
9 3 15
/ \ \
1 10 2
-----------------
Constructing binary tree
*/
tree->root = new TreeNode(4);
tree->root->left = new TreeNode(9);
tree->root->left->right = new TreeNode(5);
tree->root->left->right->left = new TreeNode(6);
tree->root->left->right->left->left = new TreeNode(9);
tree->root->left->right->right = new TreeNode(8);
tree->root->left->right->right->left = new TreeNode(3);
tree->root->left->right->right->left->left = new TreeNode(1);
tree->root->left->right->right->left->right = new TreeNode(10);
tree->root->left->left = new TreeNode(-2);
tree->root->right = new TreeNode(7);
tree->root->right->right = new TreeNode(12);
tree->root->right->right->right = new TreeNode(7);
tree->root->right->right->left = new TreeNode(5);
tree->root->right->right->left->right = new TreeNode(15);
tree->root->right->right->left->right->right = new TreeNode(2);
// Test cases
tree->leafPairSum(8);
tree->leafPairSum(2);
tree->leafPairSum(4);
tree->leafPairSum(-1);
tree->leafPairSum(9);
return 0;
}input
Given sum k : 8
(7 + 1)
Given sum k : 2
No leaf pair of given sum 2
Given sum k : 4
No leaf pair of given sum 4
Given sum k : -1
(-2 + 1)
Given sum k : 9
(7 + 2)// Include namespace system
using System;
using System.Collections.Generic;
/*
Csharp Program
Check if a pair of leaf node sum exists in given binary tree
*/
// Binary Tree node
public class TreeNode
{
public int data;
public TreeNode left;
public TreeNode right;
public TreeNode(int data)
{
// Set node value
this.data = data;
this.left = null;
this.right = null;
}
}
public class BinaryTree
{
public TreeNode root;
public BinaryTree()
{
// Set initial tree root to null
this.root = null;
}
// Find all leaf nodes using recursion
public void findLeafNodes(TreeNode node, Dictionary < int, int > record)
{
if (node == null)
{
return;
}
if (node.left == null && node.right == null)
{
if (record.ContainsKey(node.data))
{
// Increment the element frequency
record[node.data] = record[node.data] + 1;
}
else
{
// Add new leaf node
record.Add(node.data, 1);
}
}
else
{
this.findLeafNodes(node.left, record);
this.findLeafNodes(node.right, record);
}
}
// Handles the request of check pair of leaf node sum
public void leafPairSum(int k)
{
if (this.root == null)
{
// Empty Tree
return;
}
else
{
// This is use to collect leaf nodes
Dictionary < int, int > record = new Dictionary < int, int > ();
// Find all leaf nodes
this.findLeafNodes(this.root, record);
if (record.Count > 1)
{
Console.Write("\n Given sum k : " + k);
foreach(KeyValuePair < int, int > info in record)
{
// Check that pair exist in tree
if (record.ContainsKey(k - info.Key)
&& ((info.Key != k - info.Key) || info.Value > 1))
{
// Remaining amount exists in record and they are unique
// Display a pair of given sum
Console.Write("\n (" + info.Key + " + " + (k - info.Key) + ")");
return;
}
}
Console.Write("\n No leaf pair of given sum " + k);
}
else
{
Console.Write("\nSingle leaf node exists\n");
}
}
}
public static void Main(String[] args)
{
// Create new binary tree
BinaryTree tree = new BinaryTree();
/*
4
/ \
9 7
/ \ \
-2 5 12
/ \ / \
6 8 5 7
/ / \
9 3 15
/ \ \
1 10 2
-----------------
Constructing binary tree
*/
tree.root = new TreeNode(4);
tree.root.left = new TreeNode(9);
tree.root.left.right = new TreeNode(5);
tree.root.left.right.left = new TreeNode(6);
tree.root.left.right.left.left = new TreeNode(9);
tree.root.left.right.right = new TreeNode(8);
tree.root.left.right.right.left = new TreeNode(3);
tree.root.left.right.right.left.left = new TreeNode(1);
tree.root.left.right.right.left.right = new TreeNode(10);
tree.root.left.left = new TreeNode(-2);
tree.root.right = new TreeNode(7);
tree.root.right.right = new TreeNode(12);
tree.root.right.right.right = new TreeNode(7);
tree.root.right.right.left = new TreeNode(5);
tree.root.right.right.left.right = new TreeNode(15);
tree.root.right.right.left.right.right = new TreeNode(2);
// Test cases
tree.leafPairSum(8);
tree.leafPairSum(2);
tree.leafPairSum(4);
tree.leafPairSum(-1);
tree.leafPairSum(9);
}
}input
Given sum k : 8
(-2 + 10)
Given sum k : 2
No leaf pair of given sum 2
Given sum k : 4
No leaf pair of given sum 4
Given sum k : -1
(-2 + 1)
Given sum k : 9
(2 + 7)<?php
/*
Php Program
Check if a pair of leaf node sum exists in given binary tree
*/
// Binary Tree node
class TreeNode
{
public $data;
public $left;
public $right;
public function __construct($data)
{
// Set node value
$this->data = $data;
$this->left = NULL;
$this->right = NULL;
}
}
class BinaryTree
{
public $root;
public function __construct()
{
$this->root = NULL;
}
// Find all leaf nodes using recursion
public function findLeafNodes($node, &$record)
{
if ($node == NULL)
{
return;
}
if ($node->left == NULL && $node->right == NULL)
{
if (array_key_exists($node->data, $record))
{
// Increment the element frequency
$record[$node->data] = $record[$node->data] + 1;
}
else
{
// Add new leaf node
$record[$node->data] = 1;
}
}
else
{
$this->findLeafNodes($node->left, $record);
$this->findLeafNodes($node->right, $record);
}
}
// Handles the request of check pair of leaf node sum
public function leafPairSum($k)
{
if ($this->root == NULL)
{
// Empty Tree
return;
}
else
{
// This is use to collect leaf nodes
$record = array();
// Find all leaf nodes
$this->findLeafNodes($this->root, $record);
if (count($record) > 1)
{
echo("\n Given sum k : ".$k);
foreach($record as $key => $value)
{
// Check that pair exist in tree
if (array_key_exists($k - $key, $record)
&& (($key != $k - $key) || $value > 1))
{
// Remaining amount exists in record and they are unique
// Display a pair of given sum
echo ("\n (".$key." + ".($k - $key).")");
return;
}
}
echo("\n No leaf pair of given sum ".$k);
}
else
{
echo("\nSingle leaf node exists\n");
}
}
}
}
function main()
{
// Create new binary tree
$tree = new BinaryTree();
/*
4
/ \
9 7
/ \ \
-2 5 12
/ \ / \
6 8 5 7
/ / \
9 3 15
/ \ \
1 10 2
-----------------
Constructing binary tree
*/
$tree->root = new TreeNode(4);
$tree->root->left = new TreeNode(9);
$tree->root->left->right = new TreeNode(5);
$tree->root->left->right->left = new TreeNode(6);
$tree->root->left->right->left->left = new TreeNode(9);
$tree->root->left->right->right = new TreeNode(8);
$tree->root->left->right->right->left = new TreeNode(3);
$tree->root->left->right->right->left->left = new TreeNode(1);
$tree->root->left->right->right->left->right = new TreeNode(10);
$tree->root->left->left = new TreeNode(-2);
$tree->root->right = new TreeNode(7);
$tree->root->right->right = new TreeNode(12);
$tree->root->right->right->right = new TreeNode(7);
$tree->root->right->right->left = new TreeNode(5);
$tree->root->right->right->left->right = new TreeNode(15);
$tree->root->right->right->left->right->right = new TreeNode(2);
// Test cases
$tree->leafPairSum(8);
$tree->leafPairSum(2);
$tree->leafPairSum(4);
$tree->leafPairSum(-1);
$tree->leafPairSum(9);
}
main();input
Given sum k : 8
(-2 + 10)
Given sum k : 2
No leaf pair of given sum 2
Given sum k : 4
No leaf pair of given sum 4
Given sum k : -1
(-2 + 1)
Given sum k : 9
(2 + 7)/*
Node JS Program
Check if a pair of leaf node sum exists in given binary tree
*/
// Binary Tree node
class TreeNode
{
constructor(data)
{
// Set node value
this.data = data;
this.left = null;
this.right = null;
}
}
class BinaryTree
{
constructor()
{
this.root = null;
}
// Find all leaf nodes using recursion
findLeafNodes(node, record)
{
if (node == null)
{
return;
}
if (node.left == null && node.right == null)
{
if (record.has(node.data))
{
// Increment the element frequency
record.set(node.data, record.get(node.data) + 1);
}
else
{
// Add new leaf node
record.set(node.data, 1);
}
}
else
{
this.findLeafNodes(node.left, record);
this.findLeafNodes(node.right, record);
}
}
// Handles the request of check pair of leaf node sum
leafPairSum(k)
{
if (this.root == null)
{
// Empty Tree
return;
}
else
{
// This is use to collect leaf nodes
var record = new Map();
// Find all leaf nodes
this.findLeafNodes(this.root, record);
if (record.size > 1)
{
process.stdout.write("\n Given sum k : " + k);
for (let [key, value] of record)
{
// Check that pair exist in tree
if (record.has(k - key)
&& ((key != k - key) || value > 1))
{
// Remaining amount exists in record and they are unique
// Display a pair of given sum
process.stdout.write("\n (" + key + " + " + (k - key) + ")");
return;
}
}
process.stdout.write("\n No leaf pair of given sum " + k);
}
else
{
process.stdout.write("\n Single leaf node exists\n");
}
}
}
}
function main()
{
// Create new binary tree
var tree = new BinaryTree();
/*
4
/ \
9 7
/ \ \
-2 5 12
/ \ / \
6 8 5 7
/ / \
9 3 15
/ \ \
1 10 2
-----------------
Constructing binary tree
*/
tree.root = new TreeNode(4);
tree.root.left = new TreeNode(9);
tree.root.left.right = new TreeNode(5);
tree.root.left.right.left = new TreeNode(6);
tree.root.left.right.left.left = new TreeNode(9);
tree.root.left.right.right = new TreeNode(8);
tree.root.left.right.right.left = new TreeNode(3);
tree.root.left.right.right.left.left = new TreeNode(1);
tree.root.left.right.right.left.right = new TreeNode(10);
tree.root.left.left = new TreeNode(-2);
tree.root.right = new TreeNode(7);
tree.root.right.right = new TreeNode(12);
tree.root.right.right.right = new TreeNode(7);
tree.root.right.right.left = new TreeNode(5);
tree.root.right.right.left.right = new TreeNode(15);
tree.root.right.right.left.right.right = new TreeNode(2);
// Test cases
tree.leafPairSum(8);
tree.leafPairSum(2);
tree.leafPairSum(4);
tree.leafPairSum(-1);
tree.leafPairSum(9);
}
main();input
Given sum k : 8
(-2 + 10)
Given sum k : 2
No leaf pair of given sum 2
Given sum k : 4
No leaf pair of given sum 4
Given sum k : -1
(-2 + 1)
Given sum k : 9
(2 + 7)# Python 3 Program
# Check if a pair of leaf node sum exists in given binary tree
# Binary Tree node
class TreeNode :
def __init__(self, data) :
# Set node value
self.data = data
self.left = None
self.right = None
class BinaryTree :
def __init__(self) :
self.root = None
# Find all leaf nodes using recursion
def findLeafNodes(self, node, record) :
if (node == None) :
return
if (node.left == None and node.right == None) :
if (node.data in record.keys()) :
# Increment the element frequency
record[node.data] = record[node.data] + 1
else :
# Add new leaf node
record[node.data] = 1
else :
self.findLeafNodes(node.left, record)
self.findLeafNodes(node.right, record)
# Handles the request of check pair of leaf node sum
def leafPairSum(self, k) :
if (self.root == None) :
# Empty Tree
return
else :
# This is use to collect leaf nodes
record = dict()
# Find all leaf nodes
self.findLeafNodes(self.root, record)
if (len(record) > 1) :
print("\n Given sum k : ", k, end = "")
for key, value in record.items() :
# Check that pair exist in tree
if (k - key in record.keys() and
((key != k - key) or value > 1)) :
# Remaining amount exists in record and they are unique
# Display a pair of given sum
print("\n (", key ,",", (k - key) ,")", end = "")
return
print("\n No leaf pair of given sum ", k, end = "")
else :
print("\nSingle leaf node exists")
def main() :
# Create new binary tree
tree = BinaryTree()
# 4
# / \
# 9 7
# / \ \
# -2 5 12
# / \ / \
# 6 8 5 7
# / / \
# 9 3 15
# / \ \
# 1 10 2
# -----------------
# Constructing binary tree
tree.root = TreeNode(4)
tree.root.left = TreeNode(9)
tree.root.left.right = TreeNode(5)
tree.root.left.right.left = TreeNode(6)
tree.root.left.right.left.left = TreeNode(9)
tree.root.left.right.right = TreeNode(8)
tree.root.left.right.right.left = TreeNode(3)
tree.root.left.right.right.left.left = TreeNode(1)
tree.root.left.right.right.left.right = TreeNode(10)
tree.root.left.left = TreeNode(-2)
tree.root.right = TreeNode(7)
tree.root.right.right = TreeNode(12)
tree.root.right.right.right = TreeNode(7)
tree.root.right.right.left = TreeNode(5)
tree.root.right.right.left.right = TreeNode(15)
tree.root.right.right.left.right.right = TreeNode(2)
# Test cases
tree.leafPairSum(8)
tree.leafPairSum(2)
tree.leafPairSum(4)
tree.leafPairSum(-1)
tree.leafPairSum(9)
if __name__ == "__main__": main()input
Given sum k : 8
( 1 , 7 )
Given sum k : 2
No leaf pair of given sum 2
Given sum k : 4
No leaf pair of given sum 4
Given sum k : -1
( 1 , -2 )
Given sum k : 9
( 2 , 7 )# Ruby Program
# Check if a pair of leaf node sum exists in given binary tree
# Binary Tree node
class TreeNode
# Define the accessor and reader of class TreeNode
attr_reader :data, :left, :right
attr_accessor :data, :left, :right
def initialize(data)
# Set node value
self.data = data
self.left = nil
self.right = nil
end
end
class BinaryTree
# Define the accessor and reader of class BinaryTree
attr_reader :root
attr_accessor :root
def initialize()
self.root = nil
end
# Find all leaf nodes using recursion
def findLeafNodes(node, record)
if (node == nil)
return
end
if (node.left == nil && node.right == nil)
if (record.key?(node.data))
# Increment the element frequency
record[node.data] = record[node.data] + 1
else
# Add new leaf node
record[node.data] = 1
end
else
self.findLeafNodes(node.left, record)
self.findLeafNodes(node.right, record)
end
end
# Handles the request of check pair of leaf node sum
def leafPairSum(k)
if (self.root == nil)
# Empty Tree
return
else
# This is use to collect leaf nodes
record = Hash.new()
# Find all leaf nodes
self.findLeafNodes(self.root, record)
if (record.size() > 1)
print("\n Given sum k : ", k)
record.each { | key, value |
# Check that pair exist in tree
if (record.key?(k - key) &&
((key != k - key) || value > 1))
# Remaining amount exists in record and they are unique
# Display a pair of given sum
print("\n (", key ,", ", (k - key) ,")")
return
end
}
print("\n No leaf pair of given sum ", k)
else
print("\nSingle leaf node exists\n")
end
end
end
end
def main()
# Create new binary tree
tree = BinaryTree.new()
# 4
# / \
# 9 7
# / \ \
# -2 5 12
# / \ / \
# 6 8 5 7
# / / \
# 9 3 15
# / \ \
# 1 10 2
# -----------------
# Constructing binary tree
tree.root = TreeNode.new(4)
tree.root.left = TreeNode.new(9)
tree.root.left.right = TreeNode.new(5)
tree.root.left.right.left = TreeNode.new(6)
tree.root.left.right.left.left = TreeNode.new(9)
tree.root.left.right.right = TreeNode.new(8)
tree.root.left.right.right.left = TreeNode.new(3)
tree.root.left.right.right.left.left = TreeNode.new(1)
tree.root.left.right.right.left.right = TreeNode.new(10)
tree.root.left.left = TreeNode.new(-2)
tree.root.right = TreeNode.new(7)
tree.root.right.right = TreeNode.new(12)
tree.root.right.right.right = TreeNode.new(7)
tree.root.right.right.left = TreeNode.new(5)
tree.root.right.right.left.right = TreeNode.new(15)
tree.root.right.right.left.right.right = TreeNode.new(2)
# Test cases
tree.leafPairSum(8)
tree.leafPairSum(2)
tree.leafPairSum(4)
tree.leafPairSum(-1)
tree.leafPairSum(9)
end
main()input
Given sum k : 8
(-2, 10)
Given sum k : 2
No leaf pair of given sum 2
Given sum k : 4
No leaf pair of given sum 4
Given sum k : -1
(-2, 1)
Given sum k : 9
(2, 7)import scala.collection.mutable._;
/*
Scala Program
Check if a pair of leaf node sum exists in given binary tree
*/
// Binary Tree node
class TreeNode(var data: Int,
var left: TreeNode,
var right: TreeNode)
{
def this(data: Int)
{
// Set node value
this(data,null,null);
}
}
class BinaryTree(var root: TreeNode)
{
def this()
{
this(null);
}
// Find all leaf nodes using recursion
def findLeafNodes(node: TreeNode, record: HashMap[Int, Int]): Unit = {
if (node == null)
{
return;
}
if (node.left == null && node.right == null)
{
if (record.contains(node.data))
{
// Increment the element frequency
record.addOne(node.data, record.get(node.data).get + 1);
}
else
{
// Add new leaf node
record.addOne(node.data, 1);
}
}
else
{
findLeafNodes(node.left, record);
findLeafNodes(node.right, record);
}
}
// Handles the request of check pair of leaf node sum
def leafPairSum(k: Int): Unit = {
if (this.root == null)
{
// Empty Tree
return;
}
else
{
// This is use to collect leaf nodes
var record: HashMap[Int, Int] = new HashMap[Int, Int]();
// Find all leaf nodes
findLeafNodes(this.root, record);
if (record.size > 1)
{
print("\n Given sum k : " + k);
for((key, value) <- record)
{
// Check that pair exist in tree
if (record.contains(k - key) &&
((key != k - key) || value > 1))
{
// Remaining amount exists in record and they are unique
// Display a pair of given sum
print("\n (" + key + " + " + (k - key) + ")");
return;
}
}
print("\n No leaf pair of given sum " + k);
}
else
{
print("\nSingle leaf node exists\n");
}
}
}
}
object Main
{
def main(args: Array[String]): Unit = {
// Create new binary tree
var tree: BinaryTree = new BinaryTree();
/*
4
/ \
9 7
/ \ \
-2 5 12
/ \ / \
6 8 5 7
/ / \
9 3 15
/ \ \
1 10 2
-----------------
Constructing binary tree
*/
tree.root = new TreeNode(4);
tree.root.left = new TreeNode(9);
tree.root.left.right = new TreeNode(5);
tree.root.left.right.left = new TreeNode(6);
tree.root.left.right.left.left = new TreeNode(9);
tree.root.left.right.right = new TreeNode(8);
tree.root.left.right.right.left = new TreeNode(3);
tree.root.left.right.right.left.left = new TreeNode(1);
tree.root.left.right.right.left.right = new TreeNode(10);
tree.root.left.left = new TreeNode(-2);
tree.root.right = new TreeNode(7);
tree.root.right.right = new TreeNode(12);
tree.root.right.right.right = new TreeNode(7);
tree.root.right.right.left = new TreeNode(5);
tree.root.right.right.left.right = new TreeNode(15);
tree.root.right.right.left.right.right = new TreeNode(2);
// Test cases
tree.leafPairSum(8);
tree.leafPairSum(2);
tree.leafPairSum(4);
tree.leafPairSum(-1);
tree.leafPairSum(9);
}
}input
Given sum k : 8
(-2 + 10)
Given sum k : 2
No leaf pair of given sum 2
Given sum k : 4
No leaf pair of given sum 4
Given sum k : -1
(-2 + 1)
Given sum k : 9
(2 + 7)import Foundation;
/*
Swift 4 Program
Check if a pair of leaf node sum exists in given binary tree
*/
// Binary Tree node
class TreeNode
{
var data: Int;
var left: TreeNode? ;
var right: TreeNode? ;
init(_ data: Int)
{
// Set node value
self.data = data;
self.left = nil;
self.right = nil;
}
}
class BinaryTree
{
var root: TreeNode? ;
init()
{
self.root = nil;
}
// Find all leaf nodes using recursion
func findLeafNodes(_ node: TreeNode? , _ record : inout[Int : Int])
{
if (node == nil)
{
return;
}
if (node!.left == nil && node!.right == nil)
{
if (record.keys.contains(node!.data))
{
// Increment the element frequency
record[node!.data] = record[node!.data]! + 1;
}
else
{
// Add new leaf node
record[node!.data] = 1;
}
}
else
{
self.findLeafNodes(node!.left, &record);
self.findLeafNodes(node!.right, &record);
}
}
// Handles the request of check pair of leaf node sum
func leafPairSum(_ k: Int)
{
if (self.root == nil)
{
// Empty Tree
return;
}
else
{
// This is use to collect leaf nodes
var record = [Int : Int]();
// Find all leaf nodes
self.findLeafNodes(self.root, &record);
if (record.count > 1)
{
print("\n Given sum k : ", k, terminator: "");
for (key, value) in record
{
// Check that pair exist in tree
if (record.keys.contains(k - key)
&& ((key != k - key) || value > 1))
{
// Remaining amount exists in record and they are unique
// Display a pair of given sum
print("\n (", key ,"+", (k - key) ,")", terminator: "");
return;
}
}
print("\n No leaf pair of given sum ", k, terminator: "");
}
else
{
print("\nSingle leaf node exists");
}
}
}
}
func main()
{
// Create new binary tree
let tree = BinaryTree();
/*
4
/ \
9 7
/ \ \
-2 5 12
/ \ / \
6 8 5 7
/ / \
9 3 15
/ \ \
1 10 2
-----------------
Constructing binary tree
*/
tree.root = TreeNode(4);
tree.root!.left = TreeNode(9);
tree.root!.left!.right = TreeNode(5);
tree.root!.left!.right!.left = TreeNode(6);
tree.root!.left!.right!.left!.left = TreeNode(9);
tree.root!.left!.right!.right = TreeNode(8);
tree.root!.left!.right!.right!.left = TreeNode(3);
tree.root!.left!.right!.right!.left!.left = TreeNode(1);
tree.root!.left!.right!.right!.left!.right = TreeNode(10);
tree.root!.left!.left = TreeNode(-2);
tree.root!.right = TreeNode(7);
tree.root!.right!.right = TreeNode(12);
tree.root!.right!.right!.right = TreeNode(7);
tree.root!.right!.right!.left = TreeNode(5);
tree.root!.right!.right!.left!.right = TreeNode(15);
tree.root!.right!.right!.left!.right!.right = TreeNode(2);
// Test cases
tree.leafPairSum(8);
tree.leafPairSum(2);
tree.leafPairSum(4);
tree.leafPairSum(-1);
tree.leafPairSum(9);
}
main();input
Given sum k : 8
( -2 + 10 )
Given sum k : 2
No leaf pair of given sum 2
Given sum k : 4
No leaf pair of given sum 4
Given sum k : -1
( -2 + 1 )
Given sum k : 9
( 7 + 2 )/*
Kotlin Program
Check if a pair of leaf node sum exists in given binary tree
*/
// Binary Tree node
class TreeNode
{
var data: Int;
var left: TreeNode ? ;
var right: TreeNode ? ;
constructor(data: Int)
{
// Set node value
this.data = data;
this.left = null;
this.right = null;
}
}
class BinaryTree
{
var root: TreeNode ? ;
constructor()
{
this.root = null;
}
// Find all leaf nodes using recursion
fun findLeafNodes(node: TreeNode ? , record : MutableMap<Int, Int> ): Unit
{
if (node == null)
{
return;
}
if (node.left == null && node.right == null)
{
if (record.containsKey(node.data))
{
// Increment the element frequency
record.put(node.data, record.getValue(node.data) + 1);
}
else
{
// Add new leaf node
record.put(node.data, 1);
}
}
else
{
this.findLeafNodes(node.left, record);
this.findLeafNodes(node.right, record);
}
}
// Handles the request of check pair of leaf node sum
fun leafPairSum(k: Int): Unit
{
if (this.root == null)
{
// Empty Tree
return;
}
else
{
// This is use to collect leaf nodes
var record = mutableMapOf < Int , Int > ();
// Find all leaf nodes
this.findLeafNodes(this.root, record);
if (record.count() > 1)
{
print("\n Given sum k : " + k);
for ((key, value) in record)
{
// Check that pair exist in tree
if (record.containsKey(k - key)
&& ((key != k - key) || value > 1))
{
// Remaining amount exists in record and they are unique
// Display a pair of given sum
print("\n (" + key + " + " + (k - key) + ")");
return;
}
}
print("\n No leaf pair of given sum " + k);
}
else
{
print("\nSingle leaf node exists\n");
}
}
}
}
fun main(args: Array < String > ): Unit
{
// Create new binary tree
val tree: BinaryTree = BinaryTree();
/*
4
/ \
9 7
/ \ \
-2 5 12
/ \ / \
6 8 5 7
/ / \
9 3 15
/ \ \
1 10 2
-----------------
Constructing binary tree
*/
tree.root = TreeNode(4);
tree.root?.left = TreeNode(9);
tree.root?.left?.right = TreeNode(5);
tree.root?.left?.right?.left = TreeNode(6);
tree.root?.left?.right?.left?.left = TreeNode(9);
tree.root?.left?.right?.right = TreeNode(8);
tree.root?.left?.right?.right?.left = TreeNode(3);
tree.root?.left?.right?.right?.left?.left = TreeNode(1);
tree.root?.left?.right?.right?.left?.right = TreeNode(10);
tree.root?.left?.left = TreeNode(-2);
tree.root?.right = TreeNode(7);
tree.root?.right?.right = TreeNode(12);
tree.root?.right?.right?.right = TreeNode(7);
tree.root?.right?.right?.left = TreeNode(5);
tree.root?.right?.right?.left?.right = TreeNode(15);
tree.root?.right?.right?.left?.right?.right = TreeNode(2);
// Test cases
tree.leafPairSum(8);
tree.leafPairSum(2);
tree.leafPairSum(4);
tree.leafPairSum(-1);
tree.leafPairSum(9);
}input
Given sum k : 8
(-2 + 10)
Given sum k : 2
No leaf pair of given sum 2
Given sum k : 4
No leaf pair of given sum 4
Given sum k : -1
(-2 + 1)
Given sum k : 9
(2 + 7)
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