Given a Binary Search Tree and an integer sum, the task is to find all the pairs from the tree whose sum is equal to the given integer sum.
We have discussed a similar problem in this post.
Examples:
Input:
2
/ \
1 6
/ \
5 7
/
3
\
4
sum = 8
Output:
1 7
2 6
3 5
Input:
2
/ \
1 3
\
4
sum = 5
Output:
1 4
2 3
Approach: Traverse the tree in a pre-order manner from both the side, left and right, and store the values of the left and right sides into the ArrayList LeftList and RightList respectively. On reaching the leaf node, take out the left side last value and a right side last value from the respective ArrayLists. There will be three conditions:
- left side value + right side value < sum: Delete the last value of LeftList and make the left side execution to the right side because on moving from the left side to the right side in the tree the value of node increases.
- left side value + right side value > sum: Delete the last value of RightList and make the right side execution to the left side because on moving from the right side to the left side in the tree the value of node decreases.
- left side value + right side value = sum: Delete the last value of both the lists and make the left side execution to the right side and right side execution to the left side.
Below is the implementation of the above approach:
C++
// C++ implementation of the // above approach#include<bits/stdc++.h>using namespace std;struct Node{ int data; Node *left, *right, *root; Node(int data) { this -> data = data; left = NULL; right = NULL; root = NULL; }};// Function to add a node // to the BSTNode* AddNode(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 (root -> data < data) root -> right = AddNode(root -> right, data); else if (root -> data > data) root -> left = AddNode(root -> left, data); return root;}// Function to find the // target pairsvoid TargetPair(Node *node, int tar){ // LeftList which stores // the left side values vector<Node*> LeftList; // RightList which stores // the right side values vector<Node*> RightList; // curr_left pointer is used // for left side execution and // curr_right pointer is used // for right side execution Node *curr_left = node; Node *curr_right = node; while (curr_left != NULL || curr_right != NULL || LeftList.size() > 0 && RightList.size() > 0) { // Storing the left side // values into LeftList // till leaf node not found while (curr_left != NULL) { LeftList.push_back(curr_left); curr_left = curr_left -> left; } // Storing the right side // values into RightList // till leaf node not found while (curr_right != NULL) { RightList.push_back(curr_right); curr_right = curr_right -> right; } // Last node of LeftList Node *LeftNode = LeftList[LeftList.size() - 1]; // Last node of RightList Node *RightNode = RightList[RightList.size() - 1]; int leftVal = LeftNode -> data; int rightVal = RightNode -> data; // To prevent repetition // like 2, 6 and 6, 2 if (leftVal >= rightVal) break; // Delete the last value of LeftList // and make the execution to the // right side if (leftVal + rightVal < tar) { LeftList.pop_back(); curr_left = LeftNode -> right; } // Delete the last value of RightList // and make the execution to the left // side else if (leftVal + rightVal > tar) { RightList.pop_back(); curr_right = RightNode -> left; } // (left value + right value) = target // then print the left value and right value // Delete the last value of left and right list // and make the left execution to right side // and right side execution to left side else { cout << LeftNode -> data << " " << RightNode -> data << endl; RightList.pop_back(); LeftList.pop_back(); curr_left = LeftNode -> right; curr_right = RightNode -> left; } }}// Driver codeint main(){ Node *root = NULL; root = AddNode(root, 2); root = AddNode(root, 6); root = AddNode(root, 5); root = AddNode(root, 3); root = AddNode(root, 4); root = AddNode(root, 1); root = AddNode(root, 7); int sum = 8; TargetPair(root, sum);}// This code is contributed by Rutvik_56 |
Java
// Java implementation of the approachimport java.util.*;public class GFG { // A binary tree node public static class Node { int data; Node left, right, root; Node(int data) { this.data = data; } } // Function to add a node to the BST public static Node AddNode(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 (root.data < data) root.right = AddNode(root.right, data); else if (root.data > data) root.left = AddNode(root.left, data); return root; } // Function to find the target pairs public static void TargetPair(Node node, int tar) { // LeftList which stores the left side values ArrayList<Node> LeftList = new ArrayList<>(); // RightList which stores the right side values ArrayList<Node> RightList = new ArrayList<>(); // curr_left pointer is used for left side execution and // curr_right pointer is used for right side execution Node curr_left = node; Node curr_right = node; while (curr_left != null || curr_right != null || LeftList.size() > 0 && RightList.size() > 0) { // Storing the left side values into LeftList // till leaf node not found while (curr_left != null) { LeftList.add(curr_left); curr_left = curr_left.left; } // Storing the right side values into RightList // till leaf node not found while (curr_right != null) { RightList.add(curr_right); curr_right = curr_right.right; } // Last node of LeftList Node LeftNode = LeftList.get(LeftList.size() - 1); // Last node of RightList Node RightNode = RightList.get(RightList.size() - 1); int leftVal = LeftNode.data; int rightVal = RightNode.data; // To prevent repetition like 2, 6 and 6, 2 if (leftVal >= rightVal) break; // Delete the last value of LeftList and make // the execution to the right side if (leftVal + rightVal < tar) { LeftList.remove(LeftList.size() - 1); curr_left = LeftNode.right; } // Delete the last value of RightList and make // the execution to the left side else if (leftVal + rightVal > tar) { RightList.remove(RightList.size() - 1); curr_right = RightNode.left; } // (left value + right value) = target // then print the left value and right value // Delete the last value of left and right list // and make the left execution to right side // and right side execution to left side else { System.out.println(LeftNode.data + " " + RightNode.data); RightList.remove(RightList.size() - 1); LeftList.remove(LeftList.size() - 1); curr_left = LeftNode.right; curr_right = RightNode.left; } } } // Driver code public static void main(String[] b) { Node root = null; root = AddNode(root, 2); root = AddNode(root, 6); root = AddNode(root, 5); root = AddNode(root, 3); root = AddNode(root, 4); root = AddNode(root, 1); root = AddNode(root, 7); int sum = 8; TargetPair(root, sum); }} |
Python3
# Python3 implementation of the approach# A binary tree nodeclass Node: def __init__(self, key): self.data = key self.left = None self.right = None# Function to append a node to the BSTdef AddNode(root, data): # If the tree is empty, return a new node if (root == None): root = Node(data) return root # Otherwise, recur down the tree if (root.data < data): root.right = AddNode(root.right, data) elif (root.data > data): root.left = AddNode(root.left, data) return root# Function to find the target pairsdef TargetPair(node, tar): # LeftList which stores the left side values LeftList = [] # RightList which stores the right side values RightList = [] # curr_left pointer is used for left # side execution and curr_right pointer # is used for right side execution curr_left = node curr_right = node while (curr_left != None or curr_right != None or len(LeftList) > 0 and len(RightList) > 0): # Storing the left side values into # LeftList till leaf node not found while (curr_left != None): LeftList.append(curr_left) curr_left = curr_left.left # Storing the right side values into # RightList till leaf node not found while (curr_right != None): RightList.append(curr_right) curr_right = curr_right.right # Last node of LeftList LeftNode = LeftList[-1] # Last node of RightList RightNode = RightList[-1] leftVal = LeftNode.data rightVal = RightNode.data # To prevent repetition like 2, 6 and 6, 2 if (leftVal >= rightVal): break # Delete the last value of LeftList and # make the execution to the right side if (leftVal + rightVal < tar): del LeftList[-1] curr_left = LeftNode.right # Delete the last value of RightList and # make the execution to the left side elif (leftVal + rightVal > tar): del RightList[-1] curr_right = RightNode.left # (left value + right value) = target # then print the left value and right value # Delete the last value of left and right list # and make the left execution to right side # and right side execution to left side else: print(LeftNode.data, RightNode.data) del RightList[-1] del LeftList[-1] curr_left = LeftNode.right curr_right = RightNode.left# Driver codeif __name__ == '__main__': root = None root = AddNode(root, 2) root = AddNode(root, 6) root = AddNode(root, 5) root = AddNode(root, 3) root = AddNode(root, 4) root = AddNode(root, 1) root = AddNode(root, 7) sum = 8 TargetPair(root, sum)# This code is contributed by mohit kumar 29 |
C#
// C# program to implement// the above approachusing System.Collections.Generic;using System;class GFG { // A binary tree node public class Node { public int data; public Node left, right, root; public Node(int data) { this.data = data; } } // Function to add a node to the BST public static Node AddNode(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 (root.data < data) root.right = AddNode(root.right, data); else if (root.data > data) root.left = AddNode(root.left, data); return root; } // Function to find the target pairs public static void TargetPair(Node node, int tar) { // LeftList which stores the left side values List<Node> LeftList = new List<Node>(); // RightList which stores the right side values List<Node> RightList = new List<Node>(); // curr_left pointer is used for left side execution and // curr_right pointer is used for right side execution Node curr_left = node; Node curr_right = node; while (curr_left != null || curr_right != null || LeftList.Count > 0 && RightList.Count > 0) { // Storing the left side values into LeftList // till leaf node not found while (curr_left != null) { LeftList.Add(curr_left); curr_left = curr_left.left; } // Storing the right side values into RightList // till leaf node not found while (curr_right != null) { RightList.Add(curr_right); curr_right = curr_right.right; } // Last node of LeftList Node LeftNode = LeftList[LeftList.Count - 1]; // Last node of RightList Node RightNode = RightList[RightList.Count - 1]; int leftVal = LeftNode.data; int rightVal = RightNode.data; // To prevent repetition like 2, 6 and 6, 2 if (leftVal >= rightVal) break; // Delete the last value of LeftList and make // the execution to the right side if (leftVal + rightVal < tar) { LeftList.RemoveAt(LeftList.Count - 1); curr_left = LeftNode.right; } // Delete the last value of RightList and make // the execution to the left side else if (leftVal + rightVal > tar) { RightList.RemoveAt(RightList.Count - 1); curr_right = RightNode.left; } // (left value + right value) = target // then print the left value and right value // Delete the last value of left and right list // and make the left execution to right side // and right side execution to left side else { Console.WriteLine(LeftNode.data + " " + RightNode.data); RightList.RemoveAt(RightList.Count - 1); LeftList.RemoveAt(LeftList.Count - 1); curr_left = LeftNode.right; curr_right = RightNode.left; } } } // Driver code public static void Main(String[] b) { Node root = null; root = AddNode(root, 2); root = AddNode(root, 6); root = AddNode(root, 5); root = AddNode(root, 3); root = AddNode(root, 4); root = AddNode(root, 1); root = AddNode(root, 7); int sum = 8; TargetPair(root, sum); }}/* This code contributed by PrinciRaj1992 */ |
Javascript
<script>// Javascript implementation of the approach// A binary tree nodeclass Node { constructor(data) { this.data = data; this.left = this.right = null; }}// Function to add a node to the BSTfunction AddNode(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 (root.data < data) root.right = AddNode(root.right, data); else if (root.data > data) root.left = AddNode(root.left, data); return root;}// Function to find the target pairsfunction TargetPair(node, tar){ // LeftList which stores the // left side values let LeftList = []; // RightList which stores // the right side values let RightList = []; // curr_left pointer is used for left // side execution and curr_right pointer // is used for right side execution let curr_left = node; let curr_right = node; while (curr_left != null || curr_right != null || LeftList.length > 0 && RightList.length > 0) { // Storing the left side values into // LeftList till leaf node not found while (curr_left != null) { LeftList.push(curr_left); curr_left = curr_left.left; } // Storing the right side values into // RightList till leaf node not found while (curr_right != null) { RightList.push(curr_right); curr_right = curr_right.right; } // Last node of LeftList let LeftNode = LeftList[LeftList.length - 1]; // Last node of RightList let RightNode = RightList[RightList.length - 1]; let leftVal = LeftNode.data; let rightVal = RightNode.data; // To prevent repetition like 2, 6 and 6, 2 if (leftVal >= rightVal) break; // Delete the last value of LeftList and make // the execution to the right side if (leftVal + rightVal < tar) { LeftList.pop(); curr_left = LeftNode.right; } // Delete the last value of RightList and make // the execution to the left side else if (leftVal + rightVal > tar) { RightList.pop(); curr_right = RightNode.left; } // (left value + right value) = target // then print the left value and right value // Delete the last value of left and right list // and make the left execution to right side // and right side execution to left side else { document.write(LeftNode.data + " " + RightNode.data + "<br>"); RightList.pop(); LeftList.pop(); curr_left = LeftNode.right; curr_right = RightNode.left; } }}// Driver codelet root = null;root = AddNode(root, 2);root = AddNode(root, 6);root = AddNode(root, 5);root = AddNode(root, 3);root = AddNode(root, 4);root = AddNode(root, 1);root = AddNode(root, 7);let sum = 8;TargetPair(root, sum);// This code is contributed by patel2127</script> |
Output
1 7 2 6 3 5
Approach 2 using stack:
Given A BST print all the pairs with target sum present in BST. BST not contains any duplicate.
GIVEN ABOVE BST:
Input : sum = 10 Output: 0 10 1 9 2 8 3 7 4 6
Input: sum = 9 Output: 0 9 1 8 2 7 3 6 4 5
Approach discussed below is similar to find pair in sorted array using two pointer technique.
The idea used here is same as the two pointer algorithm for find pair with target sum in O(n) time
1. Create Two stacks
i) for inorder.
ii) for reverse Inorder.
2. Now populating one by one from each stack
3.
i) if sum == k we add to the sum and make find1 and find2 to false to get new elements
ii) if sum < k we add to the sum and make find1 to false.
iii) if sum == k we add to the sum and make find2 to false.
4. Breaking condition when curr1->data > curr2->data.
Below is the implementation.
C++
#include <bits/stdc++.h>using namespace std;struct TreeNode{ int data; TreeNode *right; TreeNode *left; TreeNode(int data) { this->data = data; this->right = NULL; this->left = NULL; }};TreeNode *insertNode(int data, TreeNode *root){ if (root == NULL) { TreeNode *node = new TreeNode(data); return node; } else if (data > root->data) { root->right = insertNode(data, root->right); } else if (data <= root->data) { root->left = insertNode(data, root->left); } return root;}// The idea used here is same as the two pointer algorithm for find pair with target sum in O(n) time// 1. Create Two stacks// i) for inorder// ii) for revInorder// 2. Now populating one by one from each stack// 3.// i) if sum == k we add to the sum and make find1 and find2 to false to get new elements// ii) if sum < k we add to the sum and make find1 to false.// iii) if sum == k we add to the sum and make find2 to false.// 4. breaking condition when element of curr1 > curr2void allPairs(TreeNode *root, int k){ stack<TreeNode *> s1; //inorder stack<TreeNode *> s2; // revInorder TreeNode *root1 = root, *root2 = root; TreeNode *curr1 = NULL, *curr2 = NULL; bool find1 = false, find2 = false; //markers to get new elements while (1) { // standard code for iterative inorder traversal using stack approach if (find1 == false) { while (root1 != NULL) { s1.push(root1); root1 = root1->left; } curr1 = s1.top(); s1.pop(); root1 = curr1->right; find1 = true; } // standard code for iterative reverse inorder traversal using stack approach if (find2 == false) { while (root2 != NULL) { s2.push(root2); root2 = root2->right; } curr2 = s2.top(); s2.pop(); root2 = curr2->left; find2 = true; } // breaking condition if (curr1->data >= curr2->data) { break; } // means we need next elements so make find1 and find2 to false to get next elements if (curr1->data + curr2->data == k) { cout << curr1->data << " " << curr2->data << "\n"; find1 = false; find2 = false; } // means we need greater element so make find1 to false to get next greater else if (curr1->data + curr2->data < k) { find1 = false; } // means we need smaller element so make find2 to false to get next smaller else //if (curr1->data + curr2->data > k) { find2 = false; } }}int main(){ TreeNode *root = NULL; int n = 11; int tree[] = {3, 1, 7, 0, 2, 5, 10, 4, 6, 9, 8}; for (int i = 0; i < 11; i++) { root = insertNode(tree[i], root); } allPairs(root, 10);} |
Java
import java.util.Stack;class TreeNode { int data; TreeNode right; TreeNode left; TreeNode(int data) { this.data = data; this.right = null; this.left = null; }}class BST { static TreeNode insertNode(int data, TreeNode root) { if (root == null) { TreeNode node = new TreeNode(data); return node; } else if (data > root.data) { root.right = insertNode(data, root.right); } else if (data <= root.data) { root.left = insertNode(data, root.left); } return root; } static void allPairs(TreeNode root, int k) { Stack<TreeNode> s1 = new Stack<>(); //inorder Stack<TreeNode> s2 = new Stack<>(); // revInorder TreeNode root1 = root, root2 = root; TreeNode curr1 = null, curr2 = null; boolean find1 = false, find2 = false; //markers to get new elements while (true) { // standard code for iterative inorder traversal using stack approach if (find1 == false) { while (root1 != null) { s1.push(root1); root1 = root1.left; } curr1 = s1.pop(); root1 = curr1.right; find1 = true; } // standard code for iterative reverse inorder // traversal using stack approach if (find2 == false) { while (root2 != null) { s2.push(root2); root2 = root2.right; } curr2 = s2.pop(); root2 = curr2.left; find2 = true; } // breaking condition if (curr1.data >= curr2.data) { break; } // means we need next elements so make find1 and // find2 to false to get next elements if (curr1.data + curr2.data == k) { System.out.println(curr1.data + " " + curr2.data); //System.out.println(); find1 = false; find2 = false; } // means we need greater element so make // find1 to false to get next greater else if (curr1.data + curr2.data < k) { find1 = false; } // means we need smaller element so make find2 to // false to get next smaller else { //if (curr1.data + curr2.data > k) find2 = false; } } } public static void main(String[] args) { BST bst = new BST(); TreeNode root = null; int n = 11; int[] tree = {3, 1, 7, 0, 2, 5, 10, 4, 6, 9, 8}; for (int i = 0 ; i<11; i++) root = insertNode(tree[i],root); allPairs(root,10); }} |
Python3
class TreeNode: def __init__(self,data): self.data = data self.right = None self.left = Nonedef insertNode(data, root): if (root == None): node = TreeNode(data) return node elif (data > root.data): root.right = insertNode(data, root.right) elif (data <= root.data): root.left = insertNode(data, root.left) return root# The idea used here is same as the two pointer algorithm for find pair with target sum in O(n) time# 1. Create Two stacks# i) for inorder# ii) for revInorder# 2. Now populating one by one from each stack# 3.# i) if sum == k we add to the sum and make find1 and find2 to false to get new elements# ii) if sum < k we add to the sum and make find1 to false.# iii) if sum == k we add to the sum and make find2 to false.# 4. breaking condition when element of curr1 > curr2def allPairs(root, k): s1=[] #inorder s2=[] #revInorder root1 = root; root2 = root curr1 = None; curr2 = None find1 = False; find2 = False #markers to get new elements while True: # standard code for iterative inorder traversal using stack approach if (find1 == False): while (root1 != None): s1.append(root1) root1 = root1.left curr1 = s1[-1] s1.pop() root1 = curr1.right find1 = True #standard code for iterative reverse inorder traversal using stack approach if (find2 == False): while (root2 != None): s2.append(root2) root2 = root2.right curr2 = s2[-1] s2.pop() root2 = curr2.left find2 = True # breaking condition if (curr1.data >= curr2.data): break # means we need next elements so make find1 and find2 to false to get next elements if (curr1.data + curr2.data == k): print("{} {}".format(curr1.data,curr2.data)) find1 = False find2 = False # means we need greater element so make find1 to false to get next greater elif (curr1.data + curr2.data < k): find1 = False # means we need smaller element so make find2 to false to get next smaller elif (curr1.data + curr2.data > k): find2 = Falseif __name__ == '__main__': root = None n = 11 tree = [3, 1, 7, 0, 2, 5, 10, 4, 6, 9, 8] for i in range(11): root = insertNode(tree[i], root) allPairs(root, 10) |
C#
using System;using System.Collections.Generic;public class TreeNode{ public int data; public TreeNode right; public TreeNode left; public TreeNode(int data) { this.data = data; this.right = null; this.left = null; }}public class BST{ static TreeNode insertNode(int data, TreeNode root) { if (root == null) { TreeNode node = new TreeNode(data); return node; } else if (data > root.data) { root.right = insertNode(data, root.right); } else if (data <= root.data) { root.left = insertNode(data, root.left); } return root; } static void allPairs(TreeNode root, int k) { Stack<TreeNode> s1 = new Stack<TreeNode>(); //inorder Stack<TreeNode> s2 = new Stack<TreeNode>(); // revInorder TreeNode root1 = root, root2 = root; TreeNode curr1 = null, curr2 = null; bool find1 = false, find2 = false; //markers to get new elements while (true) { // standard code for iterative inorder traversal using stack approach if (find1 == false) { while (root1 != null) { s1.Push(root1); root1 = root1.left; } curr1 = s1.Pop(); root1 = curr1.right; find1 = true; } // standard code for iterative reverse inorder // traversal using stack approach if (find2 == false) { while (root2 != null) { s2.Push(root2); root2 = root2.right; } curr2 = s2.Pop(); root2 = curr2.left; find2 = true; } // breaking condition if (curr1.data >= curr2.data) { break; } // means we need next elements so make find1 and // find2 to false to get next elements if (curr1.data + curr2.data == k) { Console.WriteLine(curr1.data + " " + curr2.data); //Console.WriteLine(); find1 = false; find2 = false; } // means we need greater element so make // find1 to false to get next greater else if (curr1.data + curr2.data < k) { find1 = false; } // means we need smaller element so make find2 to // false to get next smaller else { //if (curr1.data + curr2.data > k) find2 = false; } } } public static void Main(string[] args) { BST bst = new BST(); TreeNode root = null; int n = 11; int[] tree = { 3, 1, 7, 0, 2, 5, 10, 4, 6, 9, 8 }; for (int i = 0; i < 11; i++) root = insertNode(tree[i], root); allPairs(root, 10); }} |
Javascript
//javascript codeclass TreeNode { constructor(data) { this.data = data; this.right = null; this.left = null; }}function insertNode(data, root) { if (root == null) { node = new TreeNode(data); return node; } else if (data > root.data) { root.right = insertNode(data, root.right); } else if (data <= root.data) { root.left = insertNode(data, root.left); } return root;}function allPairs(root, k) { let s1 = []; //inorder let s2 = []; //revInorder let root1 = root; let root2 = root; let curr1 = null; let curr2 = null; let find1 = false; let find2 = false; //markers to get new elements while (true) { // standard code for iterative inorder traversal using stack approach if (find1 == false) { while (root1 != null) { s1.push(root1); root1 = root1.left; } curr1 = s1[s1.length - 1]; s1.pop(); root1 = curr1.right; find1 = true; } //standard code for iterative reverse inorder traversal using stack approach if (find2 == false) { while (root2 != null) { s2.push(root2); root2 = root2.right; } curr2 = s2[s2.length - 1]; s2.pop(); root2 = curr2.left; find2 = true; } // breaking condition if (curr1.data >= curr2.data) { break; } // means we need next elements so make find1 and find2 to false to get next elements if (curr1.data + curr2.data == k) { console.log(`${curr1.data} ${curr2.data}`); find1 = false; find2 = false; } // means we need greater element so make find1 to false to get next greater else if (curr1.data + curr2.data < k) { find1 = false; } // means we need smaller element so make find2 to false to get next smaller else if (curr1.data + curr2.data > k) { find2 = false; } }}if (true) { let root = null; let n = 11; let tree = [3, 1, 7, 0, 2, 5, 10, 4, 6, 9, 8]; for (let i = 0; i < 11; i++) { root = insertNode(tree[i], root); } allPairs(root, 10);} |
Output
0 10 1 9 2 8 3 7 4 6
Time Complexity: O(N)
Auxiliary Space: O(N)
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