Check if a Binary Tree consists of a pair of leaf nodes with sum K
Given a Binary Tree and an integer K, the task is to check if the Tree consists of a pair of leaf nodes with sum exactly K. In case of multiple pairs, print any one of them. Otherwise print -1.
Note: Assume that the given binary tree will always have more than 1 leaf node.
Examples:
Input: X = 13
1
/ \
2 3
/ \ / \
4 5 6 7
\
8
Output: [5, 8]
Explanation:
The given binary tree consists of 4 leaf nodes [4, 5, 6, 8].
The pair of nodes valued 5 and 8 have sum 13.
Input: X = 6
-1
/ \
2 3
/ \
4 5
Output: [-1]
Explanation:
The given binary tree consists of 3 leaf nodes [4, 5, 3].
No valid pair of nodes exists whose sum of their values equal to 6.
Therefore, print -1.
Naive Approach: The simplest approach to solve the problem is to traverse the tree and store all the leaf nodes in an array. Then sort the array and use two pointer technique to find if a required pair exists or not.
Time Complexity: O(NlogN)
Auxiliary Space: O(N)
Efficient Approach: The above approach can be optimized using HashSet. Follow the steps below to solve the problem:
- Create a Set to store values of leaf nodes.
- Traverse the tree and for every leaf node, check if (K – value of leaf node) exists in the unordered set or not.
- If found to be true, print the pair of node values.
- Otherwise store the value of the current node into the unordered set.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
bool pairFound = false ;
struct Node {
int data;
Node *left, *right;
};
Node* newNode( int data)
{
Node* temp = new Node();
temp->data = data;
temp->left = temp->right = NULL;
return temp;
}
void pairSum(Node* root, int target,
unordered_set< int >& S)
{
if (!root)
return ;
if (!root->left and !root->right) {
if (S.count(target - root->data)) {
cout << target - root->data << " "
<< root->data;
pairFound = true ;
return ;
}
else
S.insert(root->data);
}
pairSum(root->left, target, S);
pairSum(root->right, target, S);
}
int main()
{
Node* root = newNode(1);
root->left = newNode(2);
root->left->left = newNode(4);
root->left->right = newNode(5);
root->right = newNode(3);
root->right->left = newNode(6);
root->right->right = newNode(7);
root->right->right->right = newNode(8);
unordered_set< int > S;
int K = 13;
pairSum(root, K, S);
if (pairFound == false )
cout << "-1" ;
return 0;
}
|
Java
import java.util.*;
class GFG{
static boolean pairFound = false ;
static class Node
{
int data;
Node left, right;
};
static Node newNode( int data)
{
Node temp = new Node();
temp.data = data;
temp.left = temp.right = null ;
return temp;
}
static void pairSum(Node root, int target,
HashSet<Integer> S)
{
if (root == null )
return ;
if (root.left == null && root.right == null )
{
if (S.contains(target - root.data))
{
System.out.print(target - root.data +
" " + root.data);
pairFound = true ;
return ;
}
else
S.add(root.data);
}
pairSum(root.left, target, S);
pairSum(root.right, target, S);
}
public static void main(String[] args)
{
Node root = newNode( 1 );
root.left = newNode( 2 );
root.left.left = newNode( 4 );
root.left.right = newNode( 5 );
root.right = newNode( 3 );
root.right.left = newNode( 6 );
root.right.right = newNode( 7 );
root.right.right.right = newNode( 8 );
HashSet<Integer> S = new HashSet<Integer>();
int K = 13 ;
pairSum(root, K, S);
if (pairFound == false )
System.out.print( "-1" );
}
}
|
Python3
pairFound = False
S = set ()
class newNode:
def __init__( self , data):
self .data = data
self .left = None
self .right = None
def pairSum(root, target):
global pairFound
global S
if (root = = None ):
return
if (root.left = = None and
root.right = = None ):
temp = list (S)
if (temp.count(target - root.data)):
print (target - root.data, root.data)
pairFound = True
return
else :
S.add(root.data)
pairSum(root.left, target)
pairSum(root.right, target)
if __name__ = = '__main__' :
root = newNode( 1 )
root.left = newNode( 2 )
root.left.left = newNode( 4 )
root.left.right = newNode( 5 )
root.right = newNode( 3 )
root.right.left = newNode( 6 )
root.right.right = newNode( 7 )
root.right.right.right = newNode( 8 )
K = 13
pairSum(root, K)
if (pairFound = = False ):
print ( - 1 )
|
C#
using System;
using System.Collections.Generic;
class GFG{
static bool pairFound = false ;
class Node
{
public int data;
public Node left, right;
};
static Node newNode( int data)
{
Node temp = new Node();
temp.data = data;
temp.left = temp.right = null ;
return temp;
}
static void pairSum(Node root, int target,
HashSet< int > S)
{
if (root == null )
return ;
if (root.left == null && root.right == null )
{
if (S.Contains(target - root.data))
{
Console.Write(target - root.data +
" " + root.data);
pairFound = true ;
return ;
}
else
S.Add(root.data);
}
pairSum(root.left, target, S);
pairSum(root.right, target, S);
}
public static void Main(String[] args)
{
Node root = newNode(1);
root.left = newNode(2);
root.left.left = newNode(4);
root.left.right = newNode(5);
root.right = newNode(3);
root.right.left = newNode(6);
root.right.right = newNode(7);
root.right.right.right = newNode(8);
HashSet< int > S = new HashSet< int >();
int K = 13;
pairSum(root, K, S);
if (pairFound == false )
Console.Write( "-1" );
}
}
|
Javascript
<script>
let pairFound = false ;
class Node
{
constructor(data) {
this .left = null ;
this .right = null ;
this .data = data;
}
}
function newNode(data)
{
let temp = new Node(data);
return temp;
}
function pairSum(root, target, S)
{
if (root == null )
return ;
if (root.left == null && root.right == null )
{
if (S.has(target - root.data))
{
document.write(target - root.data + " " +
root.data);
pairFound = true ;
return ;
}
else
S.add(root.data);
}
pairSum(root.left, target, S);
pairSum(root.right, target, S);
}
let root = newNode(1);
root.left = newNode(2);
root.left.left = newNode(4);
root.left.right = newNode(5);
root.right = newNode(3);
root.right.left = newNode(6);
root.right.right = newNode(7);
root.right.right.right = newNode(8);
let S = new Set();
let K = 13;
pairSum(root, K, S);
if (pairFound == false )
document.write( "-1" );
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(N)
Last Updated :
17 Dec, 2021
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