给定两个Binary搜索树和一个整数X ，任务是找到一对节点，一个节点属于第一个BST，第二个节点属于另一个BST，以使它们的总和等于X。如果存在这样的一对，则打印“是”，否则打印“否” 。
例子：
Input: X = 100
BST 1:
5
/ \
3 7
/ \ / \
2 4 6 8
BST 2:
11
\
13
Output: No
There is no such pair with given value.
Input: X = 16
BST 1:
5
/ \
3 7
/ \ / \
2 4 6 8
BST 2:
11
\
13
Output: Yes
5 + 11 = 16
方法：我们将使用两个指针方法解决此问题。
我们将在第一个BST上创建一个正向迭代器，然后在第二个BST上创建一个反向迭代器。因此，我们将保持向前和向后的迭代器，分别按照有序遍历和反向有序遍历的顺序对BST进行迭代。
- 分别为第一个和第二个BST创建一个正向和反向迭代器。假设他们指向的节点的值是v1和v2。
- 现在在每一步
- 如果v1 + v2 = X，我们找到了一对。
- 如果v1 + v2小于或等于x，我们将使迭代器指向下一个元素。
- 如果v1 + v2大于x，我们将使反向迭代器指向上一个元素。
- 当两个迭代器都指向一个有效节点时，我们将继续上面的操作。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Node of the binary tree
struct node {
int data;
node* left;
node* right;
node(int data)
{
this->data = data;
left = NULL;
right = NULL;
}
};
// Function that returns true if a pair
// with given sum exists in the given BSTs
bool existsPair(node* root1, node* root2, int x)
{
// Stack to store nodes for forward and backward
// iterator
stack it1, it2;
// Initializing forward iterator
node* c = root1;
while (c != NULL)
it1.push(c), c = c->left;
// Initializing backward iterator
c = root2;
while (c != NULL)
it2.push(c), c = c->right;
// Two pointer technique
while (it1.size() and it2.size()) {
// To store the value of the nodes
// current iterators are pointing to
int v1 = it1.top()->data, v2 = it2.top()->data;
// If found a valid pair
if (v1 + v2 == x)
return true;
// Moving forward iterator
if (v1 + v2 < x) {
c = it1.top()->right;
it1.pop();
while (c != NULL)
it1.push(c), c = c->left;
}
// Moving backward iterator
else {
c = it2.top()->left;
it2.pop();
while (c != NULL)
it2.push(c), c = c->right;
}
}
// If no such pair found
return false;
}
// Driver code
int main()
{
// First BST
node* root1 = new node(11);
root1->right = new node(15);
// Second BST
node* root2 = new node(5);
root2->left = new node(3);
root2->right = new node(7);
root2->left->left = new node(2);
root2->left->right = new node(4);
root2->right->left = new node(6);
root2->right->right = new node(8);
int x = 23;
if (existsPair(root1, root2, x))
cout << "Yes";
else
cout << "No";
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Node of the binary tree
static class node
{
int data;
node left;
node right;
node(int data)
{
this.data = data;
left = null;
right = null;
}
};
// Function that returns true if a pair
// with given sum exists in the given BSTs
static boolean existsPair(node root1, node root2, int x)
{
// Stack to store nodes for forward and backward
// iterator
Stack it1 = new Stack(), it2 = new Stack();
// Initializing forward iterator
node c = root1;
while (c != null)
{
it1.push(c);
c = c.left;
}
// Initializing backward iterator
c = root2;
while (c != null)
{
it2.push(c);
c = c.right;
}
// Two pointer technique
while (it1.size() > 0 && it2.size() > 0)
{
// To store the value of the nodes
// current iterators are pointing to
int v1 = it1.peek().data, v2 = it2.peek().data;
// If found a valid pair
if (v1 + v2 == x)
return true;
// Moving forward iterator
if (v1 + v2 < x)
{
c = it1.peek().right;
it1.pop();
while (c != null)
{
it1.push(c); c = c.left;
}
}
// Moving backward iterator
else
{
c = it2.peek().left;
it2.pop();
while (c != null)
{
it2.push(c); c = c.right;
}
}
}
// If no such pair found
return false;
}
// Driver code
public static void main(String[] args)
{
// First BST
node root1 = new node(11);
root1.right = new node(15);
// Second BST
node root2 = new node(5);
root2.left = new node(3);
root2.right = new node(7);
root2.left.left = new node(2);
root2.left.right = new node(4);
root2.right.left = new node(6);
root2.right.right = new node(8);
int x = 23;
if (existsPair(root1, root2, x))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by Princi Singh Python3
# Python3 implementation of the approach
# Node of the binary tree
class node:
def __init__ (self, key):
self.data = key
self.left = None
self.right = None
# Function that returns true if a pair
# with given sum exists in the given BSTs
def existsPair(root1, root2, x):
# Stack to store nodes for forward
# and backward iterator
it1, it2 = [], []
# Initializing forward iterator
c = root1
while (c != None):
it1.append(c)
c = c.left
# Initializing backward iterator
c = root2
while (c != None):
it2.append(c)
c = c.right
# Two pointer technique
while (len(it1) > 0 and len(it2) > 0):
# To store the value of the nodes
# current iterators are pointing to
v1 = it1[-1].data
v2 = it2[-1].data
# If found a valid pair
if (v1 + v2 == x):
return True
# Moving forward iterator
if (v1 + v2 < x):
c = it1[-1].right
del it1[-1]
while (c != None):
it1.append(c)
c = c.left
# Moving backward iterator
else:
c = it2[-1].left
del it2[-1]
while (c != None):
it2.append(c)
c = c.right
# If no such pair found
return False
# Driver code
if __name__ == '__main__':
# First BST
root1 = node(11)
root1.right = node(15)
# Second BST
root2 = node(5)
root2.left = node(3)
root2.right = node(7)
root2.left.left = node(2)
root2.left.right = node(4)
root2.right.left = node(6)
root2.right.right = node(8)
x = 23
if (existsPair(root1, root2, x)):
print("Yes")
else:
print("No")
# This code is contributed by mohit kumar 29C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Node of the binary tree
public class node
{
public int data;
public node left;
public node right;
public node(int data)
{
this.data = data;
left = null;
right = null;
}
};
// Function that returns true if a pair
// with given sum exists in the given BSTs
static bool existsPair(node root1, node root2, int x)
{
// Stack to store nodes for forward and backward
// iterator
Stack it1 = new Stack(), it2 = new Stack();
// Initializing forward iterator
node c = root1;
while (c != null)
{
it1.Push(c);
c = c.left;
}
// Initializing backward iterator
c = root2;
while (c != null)
{
it2.Push(c);
c = c.right;
}
// Two pointer technique
while (it1.Count > 0 && it2.Count > 0)
{
// To store the value of the nodes
// current iterators are pointing to
int v1 = it1.Peek().data, v2 = it2.Peek().data;
// If found a valid pair
if (v1 + v2 == x)
return true;
// Moving forward iterator
if (v1 + v2 < x)
{
c = it1.Peek().right;
it1.Pop();
while (c != null)
{
it1.Push(c); c = c.left;
}
}
// Moving backward iterator
else
{
c = it2.Peek().left;
it2.Pop();
while (c != null)
{
it2.Push(c); c = c.right;
}
}
}
// If no such pair found
return false;
}
// Driver code
public static void Main(String[] args)
{
// First BST
node root1 = new node(11);
root1.right = new node(15);
// Second BST
node root2 = new node(5);
root2.left = new node(3);
root2.right = new node(7);
root2.left.left = new node(2);
root2.left.right = new node(4);
root2.right.left = new node(6);
root2.right.right = new node(8);
int x = 23;
if (existsPair(root1, root2, x))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by Rajput-Ji 输出：
Yes
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