Dado un árbol de búsqueda binario y una suma de enteros , la tarea es encontrar todos los pares del árbol cuya suma sea igual a la suma de enteros dada .
Hemos discutido un problema similar en esta publicación.
Ejemplos:
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
Enfoque: Atraviese el árbol de forma preordenada desde el lado izquierdo y derecho, y almacene los valores de los lados izquierdo y derecho en ArrayList LeftList y RightList respectivamente. Al llegar al Node hoja, saque el último valor del lado izquierdo y el último valor del lado derecho de las respectivas ArrayLists. Habrá tres condiciones:
- valor del lado izquierdo + valor del lado derecho <suma: Eliminar el último valor de LeftList y hacer la ejecución del lado izquierdo al lado derecho porque al moverse del lado izquierdo al lado derecho en el árbol, el valor del Node aumenta.
- valor del lado izquierdo + valor del lado derecho > suma: Eliminar el último valor de RightList y hacer la ejecución del lado derecho al lado izquierdo porque al pasar del lado derecho al lado izquierdo en el árbol el valor del Node disminuye.
- valor del lado izquierdo + valor del lado derecho = suma: elimine el último valor de ambas listas y haga la ejecución del lado izquierdo en el lado derecho y la ejecución del lado derecho en el lado izquierdo.
A continuación se muestra la implementación del enfoque anterior:
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 BST
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
void 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 code
int 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 approach
import 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 node class Node: def __init__(self, key): self.data = key self.left = None self.right = None # Function to append a node to the BST def 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 pairs def 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 code if __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 approach
using 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 node
class Node
{
constructor(data)
{
this.data = data;
this.left = this.right = null;
}
}
// Function to add a node to the BST
function 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 pairs
function 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 code
let 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>
Producción
1 7 2 6 3 5
Enfoque 2 usando la pila:
Dado A BST, imprima todos los pares con la suma objetivo presente en BST. BST no contiene ningún duplicado.
DADO POR ENCIMA DE 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
El enfoque que se analiza a continuación es similar a encontrar un par en una array ordenada utilizando la técnica de dos punteros.
La idea utilizada aquí es la misma que el algoritmo de dos punteros para encontrar un par con la suma objetivo en tiempo O (n)
1. Crea dos pilas
i) por orden.
ii) para el orden inverso.
2. Ahora llenando uno por uno de cada pila
3.
i) si sum == k sumamos a la suma y hacemos find1 y find2 a falso para obtener nuevos elementos
ii) si suma < k sumamos a la suma y hacemos que find1 sea falso.
iii) si suma == k sumamos a la suma y hacemos que find2 sea falso.
4. Condición de ruptura cuando curr1->data > curr2->data.
A continuación se muestra la implementación.
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 > curr2
void 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);
}
Python3
class TreeNode:
def __init__(self,data):
self.data = data
self.right = None
self.left = None
def 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 > curr2
def 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 = False
if __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)
Producción
0 10 1 9 2 8 3 7 4 6
Complejidad temporal: O(N)
Espacio auxiliar: O(N)