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:

Below is the implementation of the above approach:

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// C++ Program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Stores if a pair exists or not
bool pairFound = false;
 
// Struct binary tree node
struct Node {
    int data;
    Node *left, *right;
};
 
// Creates a new node
Node* newNode(int data)
{
    Node* temp = new Node();
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}
 
// Function to check if a pair of leaf
// nodes exists with sum K
void pairSum(Node* root, int target,
             unordered_set<int>& S)
{
    // checks if root is NULL
    if (!root)
        return;
 
    // Checks if the current node is a leaf node
    if (!root->left and !root->right) {
 
        // Checks for a valid pair of leaf nodes
        if (S.count(target - root->data)) {
 
            cout << target - root->data << " "
                 << root->data;
 
            pairFound = true;
            return;
        }
 
        // Insert value of current node
        // into the set
        else
            S.insert(root->data);
    }
 
    // Traverse left and right subtree
    pairSum(root->left, target, S);
    pairSum(root->right, target, S);
}
 
// Driver Code
int main()
{
    // Construct binary tree
    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);
 
    // Stores the leaf nodes
    unordered_set<int> S;
 
    int K = 13;
 
    pairSum(root, K, S);
 
    if (pairFound == false)
        cout << "-1";
 
    return 0;
}
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// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Stores if a pair exists or not
static boolean pairFound = false;
 
// Struct binary tree node
static class Node
{
    int data;
    Node left, right;
};
 
// Creates a new node
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
}
 
// Function to check if a pair of leaf
// nodes exists with sum K
static void pairSum(Node root, int target,
                    HashSet<Integer> S)
{
     
    // Checks if root is null
    if (root == null)
        return;
 
    // Checks if the current node is a leaf node
    if (root.left == null && root.right == null)
    {
         
        // Checks for a valid pair of leaf nodes
        if (S.contains(target - root.data))
        {
            System.out.print(target - root.data +
                                " " + root.data);
            pairFound = true;
            return;
        }
 
        // Insert value of current node
        // into the set
        else
            S.add(root.data);
    }
 
    // Traverse left and right subtree
    pairSum(root.left, target, S);
    pairSum(root.right, target, S);
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Construct binary tree
    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);
 
    // Stores the leaf nodes
    HashSet<Integer> S = new HashSet<Integer>();
 
    int K = 13;
 
    pairSum(root, K, S);
 
    if (pairFound == false)
        System.out.print("-1");
}
}
 
// This code is contributed by 29AjayKumar
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# Python3 program to implement
# the above approach
 
# Stores if a pair exists or not
pairFound = False
S = set()
 
# Struct binary tree node
class newNode:
     
    def __init__(self, data):
         
        self.data = data
        self.left = None
        self.right = None
 
# Function to check if a pair of
# leaf nodes exists with sum K
def pairSum(root, target):
     
    global pairFound
    global S
     
    # Checks if root is NULL
    if (root == None):
        return
 
    # Checks if the current node
    # is a leaf node
    if (root.left == None and
       root.right == None):
        temp = list(S)
         
        # Checks for a valid pair of leaf nodes
        if (temp.count(target - root.data)):
            print(target - root.data, root.data)
 
            pairFound = True
            return
         
        # Insert value of current node
        # into the set
        else:
            S.add(root.data)
 
    # Traverse left and right subtree
    pairSum(root.left, target)
    pairSum(root.right, target)
 
# Driver Code
if __name__ == '__main__':
     
    # Construct binary tree
    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)
         
# This code is contributed by bgangwar59
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// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Stores if a pair exists or not
static bool pairFound = false;
 
// Struct binary tree node
class Node
{
    public int data;
    public Node left, right;
};
 
// Creates a new node
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
}
 
// Function to check if a pair of leaf
// nodes exists with sum K
static void pairSum(Node root, int target,
                    HashSet<int> S)
{
     
    // Checks if root is null
    if (root == null)
        return;
 
    // Checks if the current node is a leaf node
    if (root.left == null && root.right == null)
    {
         
        // Checks for a valid pair of leaf nodes
        if (S.Contains(target - root.data))
        {
            Console.Write(target - root.data +
                             " " + root.data);
            pairFound = true;
            return;
        }
 
        // Insert value of current node
        // into the set
        else
            S.Add(root.data);
    }
 
    // Traverse left and right subtree
    pairSum(root.left, target, S);
    pairSum(root.right, target, S);
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Construct binary tree
    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);
 
    // Stores the leaf nodes
    HashSet<int> S = new HashSet<int>();
 
    int K = 13;
 
    pairSum(root, K, S);
 
    if (pairFound == false)
        Console.Write("-1");
}
}
 
// This code is contributed by 29AjayKumar
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Output: 
5 8


Time Complexity: O(N)
Auxiliary Space: O(N)

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