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Find a pair with given sum in a Balanced BST

  • Difficulty Level : Hard
  • Last Updated : 10 Aug, 2021

Given a Balanced Binary Search Tree and a target sum, write a function that returns true if there is a pair with sum equals to target sum, otherwise return false. Expected time complexity is O(n) and only O(Logn) extra space can be used. Any modification to Binary Search Tree is not allowed. Note that height of a Balanced BST is always O(Logn).

This problem is mainly extension of the previous post. Here we are not allowed to modify the BST.

The Brute Force Solution is to consider each pair in BST and check whether the sum equals to X. The time complexity of this solution will be O(n^2).

A Better Solution is to create an auxiliary array and store the Inorder traversal of BST in the array. The array will be sorted as Inorder traversal of BST always produces sorted data. Once we have the Inorder traversal, we can pair in O(n) time (See this for details). This solution works in O(n) time but requires O(n) auxiliary space.  



Java




// Java code to find a pair with given sum
// in a Balanced BST
import java.util.ArrayList;
 
// A binary tree node
class Node {
 
    int data;
    Node left, right;
 
    Node(int d)
    {
        data = d;
        left = right = null;
    }
}
 
class BinarySearchTree {
 
    // Root of BST
    Node root;
 
    // Constructor
    BinarySearchTree()
    {
        root = null;
    }
 
    // Inorder traversal of the tree
    void inorder()
    {
        inorderUtil(this.root);
    }
 
    // Utility function for inorder traversal of the tree
    void inorderUtil(Node node)
    {
        if (node == null)
            return;
 
        inorderUtil(node.left);
        System.out.print(node.data + " ");
        inorderUtil(node.right);
    }
 
    // This method mainly calls insertRec()
    void insert(int key)
    {
        root = insertRec(root, key);
    }
 
    /* A recursive function to insert a new key in BST */
    Node insertRec(Node root, int data)
    {
 
        /* If the tree is empty, return a new node */
        if (root == null) {
            root = new Node(data);
            return root;
        }
 
        /* Otherwise, recur down the tree */
        if (data < root.data)
            root.left = insertRec(root.left, data);
        else if (data > root.data)
            root.right = insertRec(root.right, data);
 
        return root;
    }
 
    // Method that adds values of given BST into ArrayList
    // and hence returns the ArrayList
    ArrayList<Integer> treeToList(Node node, ArrayList<Integer>
                                                 list)
    {
        // Base Case
        if (node == null)
            return list;
 
        treeToList(node.left, list);
        list.add(node.data);
        treeToList(node.right, list);
 
        return list;
    }
 
    // method that checks if there is a pair present
    boolean isPairPresent(Node node, int target)
    {
        // This list a1 is passed as an argument
        // in treeToList method
        // which is later on filled by the values of BST
        ArrayList<Integer> a1 = new ArrayList<>();
 
        // a2 list contains all the values of BST
        // returned by treeToList method
        ArrayList<Integer> a2 = treeToList(node, a1);
 
        int start = 0; // Starting index of a2
 
        int end = a2.size() - 1; // Ending index of a2
 
        while (start < end) {
 
            if (a2.get(start) + a2.get(end) == target) // Target Found!
            {
                System.out.println("Pair Found: " + a2.get(start) + " + " + a2.get(end) + " "
                                   + "= " + target);
                return true;
            }
 
            if (a2.get(start) + a2.get(end) > target) // decrements end
            {
                end--;
            }
 
            if (a2.get(start) + a2.get(end) < target) // increments start
            {
                start++;
            }
        }
 
        System.out.println("No such values are found!");
        return false;
    }
 
    // Driver function
    public static void main(String[] args)
    {
        BinarySearchTree tree = new BinarySearchTree();
        /*
                   15
                /     \
              10      20
             / \     /  \
            8  12   16  25    */
        tree.insert(15);
        tree.insert(10);
        tree.insert(20);
        tree.insert(8);
        tree.insert(12);
        tree.insert(16);
        tree.insert(25);
 
        tree.isPairPresent(tree.root, 33);
    }
}
 
// This code is contributed by Kamal Rawal

Python3




# Python3 code to find a pair with given sum
# in a Balanced BST
class Node:
     
    # Construct to create a new Node
    def __init__(self, key):
         
        self.data = key
        self.left = self.right = None
 
# A utility function to insert a new
# Node with given key in BST
def insert(root: Node, key: int):
     
    # If the tree is empty, return a new Node
    if root is None:
        return Node(key)
 
    # Otherwise, recur down the tree
    if root.data > key:
        root.left = insert(root.left, key)
 
    elif root.data < key:
        root.right = insert(root.right, key)
 
    # return the (unchanged) Node pointer
    return root
 
# Function that adds values of given BST into
# ArrayList and hence returns the ArrayList
def tree_to_list(root: Node, arr: list):
     
    if not root:
        return arr
 
    tree_to_list(root.left, arr)
    arr.append(root.data)
    tree_to_list(root.right, arr)
 
    return arr
 
# Function that checks if there is a pair present
def isPairPresent(root: Node, target: int) -> bool:
     
    # This list a1 is passed as an argument
    # in treeToList method which is later
    # on filled by the values of BST
    arr1 = []
     
    # a2 list contains all the values of BST
    # returned by treeToList method
    arr2 = tree_to_list(root, arr1)
     
    # Starting index of a2
    start = 0
     
    # Ending index of a2
    end = len(arr2) - 1
 
    while start < end:
         
        # If target found
        if arr2[start] + arr2[end] == target:
            print(f"Pair Found: {arr2[start]} + {arr2[end]} = {target}")
            return True
             
        # Decrements end
        if arr2[start] + arr2[end] > target:
            end -= 1
             
        # Increments start
        if arr2[start] + arr2[end] < target:
            start += 1
 
    print("No such values are found!")
    return False
 
# Driver code
if __name__ == "__main__":
     
    root = None
    root = insert(root, 15)
    root = insert(root, 10)
    root = insert(root, 20)
    root = insert(root, 8)
    root = insert(root, 12)
    root = insert(root, 16)
    root = insert(root, 25)
 
    isPairPresent(root, 33)
 
# This code is contributed by shindesharad71

C#




// C# code to find a pair with given sum
// in a Balanced BST
using System;
using System.Collections.Generic;
 
// A binary tree node
public class Node {
 
    public int data;
    public Node left, right;
 
    public Node(int d)
    {
        data = d;
        left = right = null;
    }
}
 
public class BinarySearchTree {
 
    // Root of BST
    Node root;
 
    // Constructor
    BinarySearchTree()
    {
        root = null;
    }
 
    // Inorder traversal of the tree
    void inorder()
    {
        inorderUtil(this.root);
    }
 
    // Utility function for inorder traversal of the tree
    void inorderUtil(Node node)
    {
        if (node == null)
            return;
 
        inorderUtil(node.left);
        Console.Write(node.data + " ");
        inorderUtil(node.right);
    }
 
    // This method mainly calls insertRec()
    void insert(int key)
    {
        root = insertRec(root, key);
    }
 
    /* A recursive function to insert a new key in BST */
    Node insertRec(Node root, int data)
    {
 
        /* If the tree is empty, return a new node */
        if (root == null) {
            root = new Node(data);
            return root;
        }
 
        /* Otherwise, recur down the tree */
        if (data < root.data)
            root.left = insertRec(root.left, data);
        else if (data > root.data)
            root.right = insertRec(root.right, data);
 
        return root;
    }
 
    // Method that adds values of given BST into ArrayList
    // and hence returns the ArrayList
    List<int> treeToList(Node node, List<int> list)
    {
        // Base Case
        if (node == null)
            return list;
 
        treeToList(node.left, list);
        list.Add(node.data);
        treeToList(node.right, list);
 
        return list;
    }
 
    // method that checks if there is a pair present
    bool isPairPresent(Node node, int target)
    {
        // This list a1 is passed as an argument
        // in treeToList method
        // which is later on filled by the values of BST
        List<int> a1 = new List<int>();
 
        // a2 list contains all the values of BST
        // returned by treeToList method
        List<int> a2 = treeToList(node, a1);
 
        int start = 0; // Starting index of a2
 
        int end = a2.Count - 1; // Ending index of a2
 
        while (start < end) {
 
            if (a2[start] + a2[end] == target) // Target Found!
            {
                Console.WriteLine("Pair Found: " + a2[start] + " + " + a2[end] + " "
                                  + "= " + target);
                return true;
            }
 
            if (a2[start] + a2[end] > target) // decrements end
            {
                end--;
            }
 
            if (a2[start] + a2[end] < target) // increments start
            {
                start++;
            }
        }
 
        Console.WriteLine("No such values are found!");
        return false;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        BinarySearchTree tree = new BinarySearchTree();
        /*
                15
                /     \
            10     20
            / \     / \
            8 12 16 25 */
        tree.insert(15);
        tree.insert(10);
        tree.insert(20);
        tree.insert(8);
        tree.insert(12);
        tree.insert(16);
        tree.insert(25);
 
        tree.isPairPresent(tree.root, 33);
    }
}
 
// This code contributed by Rajput-Ji

Javascript




<script>
 
      // JavaScript code to find a pair with given sum
      // in a Balanced BST
      // A binary tree node
      class Node {
        constructor(d) {
          this.data = d;
          this.left = null;
          this.right = null;
        }
      }
 
      class BinarySearchTree {
        // Constructor
        constructor() {
          this.root = null;
        }
 
        // Inorder traversal of the tree
        inorder() {
          this.inorderUtil(this.root);
        }
 
        // Utility function for inorder traversal of the tree
        inorderUtil(node) {
          if (node == null) return;
 
          this.inorderUtil(node.left);
          document.write(node.data + " ");
          this.inorderUtil(node.right);
        }
 
        // This method mainly calls insertRec()
        insert(key) {
          this.root = this.insertRec(this.root, key);
        }
 
        /* A recursive function to insert a new key in BST */
        insertRec(root, data) {
          /* If the tree is empty, return a new node */
          if (root == null) {
            root = new Node(data);
            return root;
          }
 
          /* Otherwise, recur down the tree */
          if (data < root.data)
          root.left = this.insertRec(root.left, data);
          else if (data > root.data)
            root.right = this.insertRec(root.right, data);
 
          return root;
        }
 
        // Method that adds values of given BST into ArrayList
        // and hence returns the ArrayList
        treeToList(node, list) {
          // Base Case
          if (node == null) return list;
 
          this.treeToList(node.left, list);
          list.push(node.data);
          this.treeToList(node.right, list);
 
          return list;
        }
 
        // method that checks if there is a pair present
        isPairPresent(node, target) {
          // This list a1 is passed as an argument
          // in treeToList method
          // which is later on filled by the values of BST
          var a1 = [];
 
          // a2 list contains all the values of BST
          // returned by treeToList method
          var a2 = this.treeToList(node, a1);
 
          var start = 0; // Starting index of a2
 
          var end = a2.length - 1; // Ending index of a2
 
          while (start < end) {
            if (a2[start] + a2[end] == target) {
              // Target Found!
              document.write(
                "Pair Found: " +
                  a2[start] +
                  " + " +
                  a2[end] +
                  " " +
                  "= " +
                  target +
                  "<br>"
              );
              return true;
            }
 
            if (a2[start] + a2[end] > target) {
              // decrements end
              end--;
            }
 
            if (a2[start] + a2[end] < target) {
              // increments start
              start++;
            }
          }
 
          document.write("No such values are found!");
          return false;
        }
      }
      // Driver code
      var tree = new BinarySearchTree();
      /*
                15
                /     \
            10     20
            / \     / \
            8 12 16 25 */
      tree.insert(15);
      tree.insert(10);
      tree.insert(20);
      tree.insert(8);
      tree.insert(12);
      tree.insert(16);
      tree.insert(25);
 
      tree.isPairPresent(tree.root, 33);
       
</script>

Output :  

Pair Found: 8 + 25 = 33
  • Complexity Analysis: 
    • Time Complexity: O(n). 
      Inorder Traversal of BST takes linear time.
    • Auxiliary Space: O(n). 
      Use of array for storing the Inorder Traversal.

A space optimized solution is discussed in previous post. The idea was to first in-place convert BST to Doubly Linked List (DLL), then find pair in sorted DLL in O(n) time. This solution takes O(n) time and O(Logn) extra space, but it modifies the given BST.

The solution discussed below takes O(n) time, O(Logn) space and doesn’t modify BST. The idea is same as finding the pair in sorted array (See method 1 of this for details). We traverse BST in Normal Inorder and Reverse Inorder simultaneously. In reverse inorder, we start from the rightmost node which is the maximum value node. In normal inorder, we start from the left most node which is minimum value node. We add sum of current nodes in both traversals and compare this sum with given target sum. If the sum is same as target sum, we return true. If the sum is more than target sum, we move to next node in reverse inorder traversal, otherwise we move to next node in normal inorder traversal. If any of the traversals is finished without finding a pair, we return false.

Following is the implementation of this approach. 

C++




/* In a balanced binary search tree
isPairPresent two element which sums to
a given value time O(n) space O(logn) */
#include <bits/stdc++.h>
using namespace std;
#define MAX_SIZE 100
 
// A BST node
class node {
public:
    int val;
    node *left, *right;
};
 
// Stack type
class Stack {
public:
    int size;
    int top;
    node** array;
};
 
// A utility function to create a stack of given size
Stack* createStack(int size)
{
    Stack* stack = new Stack();
    stack->size = size;
    stack->top = -1;
    stack->array = new node*[(stack->size * sizeof(node*))];
    return stack;
}
 
// BASIC OPERATIONS OF STACK
int isFull(Stack* stack)
{
    return stack->top - 1 == stack->size;
}
 
int isEmpty(Stack* stack)
{
    return stack->top == -1;
}
 
void push(Stack* stack, node* node)
{
    if (isFull(stack))
        return;
    stack->array[++stack->top] = node;
}
 
node* pop(Stack* stack)
{
    if (isEmpty(stack))
        return NULL;
    return stack->array[stack->top--];
}
 
// Returns true if a pair with target
// sum exists in BST, otherwise false
bool isPairPresent(node* root, int target)
{
    // Create two stacks. s1 is used for
    // normal inorder traversal and s2 is
    // used for reverse inorder traversal
    Stack* s1 = createStack(MAX_SIZE);
    Stack* s2 = createStack(MAX_SIZE);
 
    // Note the sizes of stacks is MAX_SIZE,
    // we can find the tree size and fix stack size
    // as O(Logn) for balanced trees like AVL and Red Black
    // tree. We have used MAX_SIZE to keep the code simple
 
    // done1, val1 and curr1 are used for
    // normal inorder traversal using s1
    // done2, val2 and curr2 are used for
    // reverse inorder traversal using s2
    bool done1 = false, done2 = false;
    int val1 = 0, val2 = 0;
    node *curr1 = root, *curr2 = root;
 
    // The loop will break when we either find a pair or one of the two
    // traversals is complete
    while (1) {
        // Find next node in normal Inorder
        // traversal. See following post
        // https:// www.geeksforgeeks.org/inorder-tree-traversal-without-recursion/
        while (done1 == false) {
            if (curr1 != NULL) {
                push(s1, curr1);
                curr1 = curr1->left;
            }
            else {
                if (isEmpty(s1))
                    done1 = 1;
                else {
                    curr1 = pop(s1);
                    val1 = curr1->val;
                    curr1 = curr1->right;
                    done1 = 1;
                }
            }
        }
 
        // Find next node in REVERSE Inorder traversal. The only
        // difference between above and below loop is, in below loop
        // right subtree is traversed before left subtree
        while (done2 == false) {
            if (curr2 != NULL) {
                push(s2, curr2);
                curr2 = curr2->right;
            }
            else {
                if (isEmpty(s2))
                    done2 = 1;
                else {
                    curr2 = pop(s2);
                    val2 = curr2->val;
                    curr2 = curr2->left;
                    done2 = 1;
                }
            }
        }
 
        // If we find a pair, then print the pair and return. The first
        // condition makes sure that two same values are not added
        if ((val1 != val2) && (val1 + val2) == target) {
            cout << "Pair Found: " << val1 << "+ " << val2 << " = " << target << endl;
            return true;
        }
 
        // If sum of current values is smaller,
        // then move to next node in
        // normal inorder traversal
        else if ((val1 + val2) < target)
            done1 = false;
 
        // If sum of current values is greater,
        // then move to next node in
        // reverse inorder traversal
        else if ((val1 + val2) > target)
            done2 = false;
 
        // If any of the inorder traversals is
        // over, then there is no pair
        // so return false
        if (val1 >= val2)
            return false;
    }
}
 
// A utility function to create BST node
node* NewNode(int val)
{
    node* tmp = new node();
    tmp->val = val;
    tmp->right = tmp->left = NULL;
    return tmp;
}
 
// Driver program to test above functions
int main()
{
    /*
                15
                / \
            10 20
            / \ / \
            8 12 16 25 */
    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);
 
    int target = 33;
    if (isPairPresent(root, target) == false)
        cout << "\nNo such values are found\n";
    return 0;
}
 
// This code is contributed by rathbhupendra

C




/* In a balanced binary search tree isPairPresent two element which sums to
   a given value time O(n) space O(logn) */
#include <stdio.h>
#include <stdlib.h>
#define MAX_SIZE 100
 
// A BST node
struct node {
    int val;
    struct node *left, *right;
};
 
// Stack type
struct Stack {
    int size;
    int top;
    struct node** array;
};
 
// A utility function to create a stack of given size
struct Stack* createStack(int size)
{
    struct Stack* stack = (struct Stack*)malloc(sizeof(struct Stack));
    stack->size = size;
    stack->top = -1;
    stack->array = (struct node**)malloc(stack->size * sizeof(struct node*));
    return stack;
}
 
// BASIC OPERATIONS OF STACK
int isFull(struct Stack* stack)
{
    return stack->top - 1 == stack->size;
}
 
int isEmpty(struct Stack* stack)
{
    return stack->top == -1;
}
 
void push(struct Stack* stack, struct node* node)
{
    if (isFull(stack))
        return;
    stack->array[++stack->top] = node;
}
 
struct node* pop(struct Stack* stack)
{
    if (isEmpty(stack))
        return NULL;
    return stack->array[stack->top--];
}
 
// Returns true if a pair with target sum exists in BST, otherwise false
bool isPairPresent(struct node* root, int target)
{
    // Create two stacks. s1 is used for normal inorder traversal
    // and s2 is used for reverse inorder traversal
    struct Stack* s1 = createStack(MAX_SIZE);
    struct Stack* s2 = createStack(MAX_SIZE);
 
    // Note the sizes of stacks is MAX_SIZE, we can find the tree size and
    // fix stack size as O(Logn) for balanced trees like AVL and Red Black
    // tree. We have used MAX_SIZE to keep the code simple
 
    // done1, val1 and curr1 are used for normal inorder traversal using s1
    // done2, val2 and curr2 are used for reverse inorder traversal using s2
    bool done1 = false, done2 = false;
    int val1 = 0, val2 = 0;
    struct node *curr1 = root, *curr2 = root;
 
    // The loop will break when we either find a pair or one of the two
    // traversals is complete
    while (1) {
        // Find next node in normal Inorder traversal. See following post
        // https:// www.geeksforgeeks.org/inorder-tree-traversal-without-recursion/
        while (done1 == false) {
            if (curr1 != NULL) {
                push(s1, curr1);
                curr1 = curr1->left;
            }
            else {
                if (isEmpty(s1))
                    done1 = 1;
                else {
                    curr1 = pop(s1);
                    val1 = curr1->val;
                    curr1 = curr1->right;
                    done1 = 1;
                }
            }
        }
 
        // Find next node in REVERSE Inorder traversal. The only
        // difference between above and below loop is, in below loop
        // right subtree is traversed before left subtree
        while (done2 == false) {
            if (curr2 != NULL) {
                push(s2, curr2);
                curr2 = curr2->right;
            }
            else {
                if (isEmpty(s2))
                    done2 = 1;
                else {
                    curr2 = pop(s2);
                    val2 = curr2->val;
                    curr2 = curr2->left;
                    done2 = 1;
                }
            }
        }
 
        // If we find a pair, then print the pair and return. The first
        // condition makes sure that two same values are not added
        if ((val1 != val2) && (val1 + val2) == target) {
            printf("\n Pair Found: %d + %d = %d\n", val1, val2, target);
            return true;
        }
 
        // If sum of current values is smaller, then move to next node in
        // normal inorder traversal
        else if ((val1 + val2) < target)
            done1 = false;
 
        // If sum of current values is greater, then move to next node in
        // reverse inorder traversal
        else if ((val1 + val2) > target)
            done2 = false;
 
        // If any of the inorder traversals is over, then there is no pair
        // so return false
        if (val1 >= val2)
            return false;
    }
}
 
// A utility function to create BST node
struct node* NewNode(int val)
{
    struct node* tmp = (struct node*)malloc(sizeof(struct node));
    tmp->val = val;
    tmp->right = tmp->left = NULL;
    return tmp;
}
 
// Driver program to test above functions
int main()
{
    /*
                   15
                /     \
              10      20
             / \     /  \
            8  12   16  25    */
    struct 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);
 
    int target = 33;
    if (isPairPresent(root, target) == false)
        printf("\n No such values are found\n");
 
    getchar();
    return 0;
}

Java




/* In a balanced binary search tree
isPairPresent two element which sums to
a given value time O(n) space O(logn) */
import java.util.*;
class GFG
{
static final int MAX_SIZE= 100;
 
// A BST node
static class node
{
    int val;
    node left, right;
};
 
// Stack type
static class Stack
{
    int size;
    int top;
    node []array;
};
 
// A utility function to create a stack of given size
static Stack createStack(int size)
{
    Stack stack = new Stack();
    stack.size = size;
    stack.top = -1;
    stack.array = new node[stack.size];
    return stack;
}
 
// BASIC OPERATIONS OF STACK
static int isFull(Stack stack)
{
    return (stack.top - 1 == stack.size)?1:0 ;
}
 
static int isEmpty(Stack stack)
{
    return stack.top == -1?1:0;
}
 
static void push(Stack stack, node node)
{
    if (isFull(stack)==1)
        return;
    stack.array[++stack.top] = node;
}
 
static node pop(Stack stack)
{
    if (isEmpty(stack) == 1)
        return null;
    return stack.array[stack.top--];
}
 
// Returns true if a pair with target
// sum exists in BST, otherwise false
static boolean isPairPresent(node root, int target)
{
    // Create two stacks. s1 is used for
    // normal inorder traversal and s2 is
    // used for reverse inorder traversal
    Stack s1 = createStack(MAX_SIZE);
    Stack s2 = createStack(MAX_SIZE);
 
    // Note the sizes of stacks is MAX_SIZE,
    // we can find the tree size and fix stack size
    // as O(Logn) for balanced trees like AVL and Red Black
    // tree. We have used MAX_SIZE to keep the code simple
 
    // done1, val1 and curr1 are used for
    // normal inorder traversal using s1
    // done2, val2 and curr2 are used for
    // reverse inorder traversal using s2
    boolean done1 = false, done2 = false;
    int val1 = 0, val2 = 0;
    node curr1 = root, curr2 = root;
 
    // The loop will break when we either
  // find a pair or one of the two
    // traversals is complete
    while (true)
    {
       
        // Find next node in normal Inorder
        // traversal. See following post
        // https:// www.geeksforgeeks.org/inorder-tree-traversal-without-recursion/
        while (done1 == false)
        {
            if (curr1 != null)
            {
                push(s1, curr1);
                curr1 = curr1.left;
            }
            else
            {
                if (isEmpty(s1) == 1)
                    done1 = true;
                else
                {
                    curr1 = pop(s1);
                    val1 = curr1.val;
                    curr1 = curr1.right;
                    done1 = true;
                }
            }
        }
 
        // Find next node in REVERSE Inorder traversal. The only
        // difference between above and below loop is, in below loop
        // right subtree is traversed before left subtree
        while (done2 == false)
        {
            if (curr2 != null)
            {
                push(s2, curr2);
                curr2 = curr2.right;
            }
            else {
                if (isEmpty(s2) == 1)
                    done2 = true;
                else {
                    curr2 = pop(s2);
                    val2 = curr2.val;
                    curr2 = curr2.left;
                    done2 = true;
                }
            }
        }
 
        // If we find a pair, then print the pair and return. The first
        // condition makes sure that two same values are not added
        if ((val1 != val2) && (val1 + val2) == target)
        {
            System.out.print("Pair Found: " +
                             val1+ "+ " +
                             val2+ " = "
                             target +"\n");
            return true;
        }
 
        // If sum of current values is smaller,
        // then move to next node in
        // normal inorder traversal
        else if ((val1 + val2) < target)
            done1 = false;
 
        // If sum of current values is greater,
        // then move to next node in
        // reverse inorder traversal
        else if ((val1 + val2) > target)
            done2 = false;
 
        // If any of the inorder traversals is
        // over, then there is no pair
        // so return false
        if (val1 >= val2)
            return false;
    }
}
 
// A utility function to create BST node
static node NewNode(int val)
{
    node tmp = new node();
    tmp.val = val;
    tmp.right = tmp.left = null;
    return tmp;
}
 
// Driver program to test above functions
public static void main(String[] args)
{
    /*
                15
                / \
            10 20
            / \ / \
            8 12 16 25 */
    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);
 
    int target = 33;
    if (isPairPresent(root, target) == false)
        System.out.print("\nNo such values are found\n");
}
}
 
// This code is contributed by aashish1995

C#




/* In a balanced binary search tree
isPairPresent two element which sums to
a given value time O(n) space O(logn) */
using System;
using System.Collections.Generic;
 
public class GFG
{
static readonly int MAX_SIZE= 100;
 
// A BST node
public
 class node
{
    public
 int val;
    public
 node left, right;
};
 
// Stack type
public
 class Stack
{
    public
 int size;
    public
 int top;
    public
 node []array;
};
 
// A utility function to create a stack of given size
static Stack createStack(int size)
{
    Stack stack = new Stack();
    stack.size = size;
    stack.top = -1;
    stack.array = new node[stack.size];
    return stack;
}
 
// BASIC OPERATIONS OF STACK
static int isFull(Stack stack)
{
    return (stack.top - 1 == stack.size) ? 1 : 0 ;
}
 
static int isEmpty(Stack stack)
{
    return stack.top == -1?1:0;
}
 
static void push(Stack stack, node node)
{
    if (isFull(stack)==1)
        return;
    stack.array[++stack.top] = node;
}
 
static node pop(Stack stack)
{
    if (isEmpty(stack) == 1)
        return null;
    return stack.array[stack.top--];
}
 
// Returns true if a pair with target
// sum exists in BST, otherwise false
static bool isPairPresent(node root, int target)
{
    // Create two stacks. s1 is used for
    // normal inorder traversal and s2 is
    // used for reverse inorder traversal
    Stack s1 = createStack(MAX_SIZE);
    Stack s2 = createStack(MAX_SIZE);
 
    // Note the sizes of stacks is MAX_SIZE,
    // we can find the tree size and fix stack size
    // as O(Logn) for balanced trees like AVL and Red Black
    // tree. We have used MAX_SIZE to keep the code simple
 
    // done1, val1 and curr1 are used for
    // normal inorder traversal using s1
    // done2, val2 and curr2 are used for
    // reverse inorder traversal using s2
    bool done1 = false, done2 = false;
    int val1 = 0, val2 = 0;
    node curr1 = root, curr2 = root;
 
    // The loop will break when we either
  // find a pair or one of the two
    // traversals is complete
    while (true)
    {
       
        // Find next node in normal Inorder
        // traversal. See following post
        // https:// www.geeksforgeeks.org/inorder-tree-traversal-without-recursion/
        while (done1 == false)
        {
            if (curr1 != null)
            {
                push(s1, curr1);
                curr1 = curr1.left;
            }
            else
            {
                if (isEmpty(s1) == 1)
                    done1 = true;
                else
                {
                    curr1 = pop(s1);
                    val1 = curr1.val;
                    curr1 = curr1.right;
                    done1 = true;
                }
            }
        }
 
        // Find next node in REVERSE Inorder traversal. The only
        // difference between above and below loop is, in below loop
        // right subtree is traversed before left subtree
        while (done2 == false)
        {
            if (curr2 != null)
            {
                push(s2, curr2);
                curr2 = curr2.right;
            }
            else {
                if (isEmpty(s2) == 1)
                    done2 = true;
                else {
                    curr2 = pop(s2);
                    val2 = curr2.val;
                    curr2 = curr2.left;
                    done2 = true;
                }
            }
        }
 
        // If we find a pair, then print the pair and return. The first
        // condition makes sure that two same values are not added
        if ((val1 != val2) && (val1 + val2) == target)
        {
            Console.Write("Pair Found: " +
                             val1+ "+ " +
                             val2+ " = "
                             target +"\n");
            return true;
        }
 
        // If sum of current values is smaller,
        // then move to next node in
        // normal inorder traversal
        else if ((val1 + val2) < target)
            done1 = false;
 
        // If sum of current values is greater,
        // then move to next node in
        // reverse inorder traversal
        else if ((val1 + val2) > target)
            done2 = false;
 
        // If any of the inorder traversals is
        // over, then there is no pair
        // so return false
        if (val1 >= val2)
            return false;
    }
}
 
// A utility function to create BST node
static node NewNode(int val)
{
    node tmp = new node();
    tmp.val = val;
    tmp.right = tmp.left = null;
    return tmp;
}
 
// Driver program to test above functions
public static void Main(String[] args)
{
    /*
                15
                / \
            10 20
            / \ / \
            8 12 16 25 */
    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);
 
    int target = 33;
    if (isPairPresent(root, target) == false)
        Console.Write("\nNo such values are found\n");
}
}
 
// This code is contributed by aashish1995

Javascript




<script>
/* In a balanced binary search tree
isPairPresent two element which sums to
a given value time O(n) space O(logn) */
 
let MAX_SIZE= 100;
 
// A BST node
class Node
{
    constructor(val)
    {
        this.val = val;
        this.left = this.right = null;
    }
}
 
// Stack type
class Stack
{
    constructor()
    {
        this.size = 0;
        this.top = 0;
        this.array;
    }
}
 
// A utility function to create a stack of given size
function createStack(size)
{
    let stack = new Stack();
    stack.size = size;
    stack.top = -1;
    stack.array = new Array(stack.size);
    return stack;
}
 
// BASIC OPERATIONS OF STACK
function isFull(stack)
{
    return (stack.top - 1 == stack.size)?1:0 ;
}
 
function isEmpty(stack)
{
    return stack.top == -1?1:0;
}
 
function push(stack,node)
{
    if (isFull(stack)==1)
        return;
    stack.array[++stack.top] = node;
}
 
function pop(stack)
{
    if (isEmpty(stack) == 1)
        return null;
    return stack.array[stack.top--];
}
 
// Returns true if a pair with target
// sum exists in BST, otherwise false
function isPairPresent(root,target)
{
    // Create two stacks. s1 is used for
    // normal inorder traversal and s2 is
    // used for reverse inorder traversal
    let s1 = createStack(MAX_SIZE);
    let s2 = createStack(MAX_SIZE);
  
    // Note the sizes of stacks is MAX_SIZE,
    // we can find the tree size and fix stack size
    // as O(Logn) for balanced trees like AVL and Red Black
    // tree. We have used MAX_SIZE to keep the code simple
  
    // done1, val1 and curr1 are used for
    // normal inorder traversal using s1
    // done2, val2 and curr2 are used for
    // reverse inorder traversal using s2
    let done1 = false, done2 = false;
    let val1 = 0, val2 = 0;
    let curr1 = root, curr2 = root;
  
    // The loop will break when we either
  // find a pair or one of the two
    // traversals is complete
    while (true)
    {
        
        // Find next node in normal Inorder
        // traversal. See following post
        // https:// www.geeksforgeeks.org/inorder-tree-traversal-without-recursion/
        while (done1 == false)
        {
            if (curr1 != null)
            {
                push(s1, curr1);
                curr1 = curr1.left;
            }
            else
            {
                if (isEmpty(s1) == 1)
                    done1 = true;
                else
                {
                    curr1 = pop(s1);
                    val1 = curr1.val;
                    curr1 = curr1.right;
                    done1 = true;
                }
            }
        }
  
        // Find next node in REVERSE Inorder traversal. The only
        // difference between above and below loop is, in below loop
        // right subtree is traversed before left subtree
        while (done2 == false)
        {
            if (curr2 != null)
            {
                push(s2, curr2);
                curr2 = curr2.right;
            }
            else {
                if (isEmpty(s2) == 1)
                    done2 = true;
                else {
                    curr2 = pop(s2);
                    val2 = curr2.val;
                    curr2 = curr2.left;
                    done2 = true;
                }
            }
        }
  
        // If we find a pair, then print the pair and return. The first
        // condition makes sure that two same values are not added
        if ((val1 != val2) && (val1 + val2) == target)
        {
            document.write("Pair Found: " +
                             val1+ "+ " +
                             val2+ " = " +
                             target +"<br>");
            return true;
        }
  
        // If sum of current values is smaller,
        // then move to next node in
        // normal inorder traversal
        else if ((val1 + val2) < target)
            done1 = false;
  
        // If sum of current values is greater,
        // then move to next node in
        // reverse inorder traversal
        else if ((val1 + val2) > target)
            done2 = false;
  
        // If any of the inorder traversals is
        // over, then there is no pair
        // so return false
        if (val1 >= val2)
            return false;
    }
}
 
// Driver program to test above functions
/*
                15
                / \
            10 20
            / \ / \
            8 12 16 25 */
let root = new Node(15);
root.left = new Node(10);
root.right = new Node(20);
root.left.left = new Node(8);
root.left.right = new Node(12);
root.right.left = new Node(16);
root.right.right = new Node(25);
 
let target = 33;
if (isPairPresent(root, target) == false)
    document.write("<br>No such values are found<br>");
     
// This code is contributed by avanitrachhadiya2155
</script>

Output: 

 Pair Found: 8 + 25 = 33
  • Complexity Analysis: 
    • Time Complexity: O(n). 
      Inorder Traversal of BST takes linear time.
    • Auxiliary Space: O(logn). 
      The stack holds log N values as at a single time

 

https://www.youtube.com/embed/TvAFvAoS6s8?list=PLqM7alHXFySHCXD7r1J0ky9Zg_GBB1dbk
 

This article is compiled by Kumar and reviewed by GeeksforGeeks team. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

 

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