给定一个二分搜索树和一个整数X ，任务是检查BST中是否存在一对总和等于X的不同节点。如果是，则打印“是”，否则打印“否” 。
例子：
Input: X = 5
5
/ \
3 7
/ \ / \
2 4 6 8
Output: Yes
2 + 3 = 5. Thus, the answer is "Yes"
Input: X = 10
1
\
2
\
3
\
4
\
5
Output: No
方法：我们已经在本文中讨论了基于哈希的方法。其空间复杂度为O(N)，其中N是BST中的节点数。
在本文中，我们将通过节省空间的方法将空间复杂度降低到O(H)(其中H是BST的高度)来使用空间高效的方法解决相同的问题。为此，我们将在BST上使用两个指针技术。因此，我们将维护一个向前和向后的迭代器，分别以有序遍历和反向有序遍历的顺序对BST进行迭代。以下是解决问题的步骤：
- 为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
- 如果v1 + v2> x，我们将使反向迭代器指向上一个元素。
- 如果我们找不到这样的一对，答案将为“否”。
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 to find a pair with given sum
bool existsPair(node* root, int x)
{
// Iterators for BST
stack it1, it2;
// Initializing forward iterator
node* c = root;
while (c != NULL)
it1.push(c), c = c->left;
// Initializing backward iterator
c = root;
while (c != NULL)
it2.push(c), c = c->right;
// Two pointer technique
while (it1.top() != it2.top()) {
// Variables to store values at
// it1 and it2
int v1 = it1.top()->data, v2 = it2.top()->data;
// Base case
if (v1 + v2 == x)
return true;
// Moving forward pointer
if (v1 + v2 < x) {
c = it1.top()->right;
it1.pop();
while (c != NULL)
it1.push(c), c = c->left;
}
// Moving backward pointer
else {
c = it2.top()->left;
it2.pop();
while (c != NULL)
it2.push(c), c = c->right;
}
}
// Case when no pair is found
return false;
}
// Driver code
int main()
{
node* root = new node(5);
root->left = new node(3);
root->right = new node(7);
root->left->left = new node(2);
root->left->right = new node(4);
root->right->left = new node(6);
root->right->right = new node(8);
int x = 5;
// Calling required function
if (existsPair(root, 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 to find a pair with given sum
static boolean existsPair(node root, int x)
{
// Iterators for BST
Stack it1 = new Stack(), it2 = new Stack();
// Initializing forward iterator
node c = root;
while (c != null)
{
it1.push(c);
c = c.left;
}
// Initializing backward iterator
c = root;
while (c != null)
{
it2.push(c);
c = c.right;
}
// Two pointer technique
while (it1.peek() != it2.peek())
{
// Variables to store values at
// it1 and it2
int v1 = it1.peek().data, v2 = it2.peek().data;
// Base case
if (v1 + v2 == x)
return true;
// Moving forward pointer
if (v1 + v2 < x)
{
c = it1.peek().right;
it1.pop();
while (c != null)
{
it1.push(c);
c = c.left;
}
}
// Moving backward pointer
else
{
c = it2.peek().left;
it2.pop();
while (c != null)
{
it2.push(c);
c = c.right;
}
}
}
// Case when no pair is found
return false;
}
// Driver code
public static void main(String[] args)
{
node root = new node(5);
root.left = new node(3);
root.right = new node(7);
root.left.left = new node(2);
root.left.right = new node(4);
root.right.left = new node(6);
root.right.right = new node(8);
int x = 5;
// Calling required function
if (existsPair(root, x))
System.out.print("Yes");
else
System.out.print("No");
}
}
// This code is contributed by 29AjayKumar 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, 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 = root1
while (c != None):
it2.append(c)
c = c.right
# Two pointer technique
while (it1[-1] != it2[-1]):
# To store the value of the nodes
# current iterators are pointing to
v1 = it1[-1].data
v2 = it2[-1].data
# Base case
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__':
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 = 5
# Calling required function
if (existsPair(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 to find a pair with given sum
static bool existsPair(node root, int x)
{
// Iterators for BST
Stack it1 = new Stack(),
it2 = new Stack();
// Initializing forward iterator
node c = root;
while (c != null)
{
it1.Push(c);
c = c.left;
}
// Initializing backward iterator
c = root;
while (c != null)
{
it2.Push(c);
c = c.right;
}
// Two pointer technique
while (it1.Peek() != it2.Peek())
{
// Variables to store values at
// it1 and it2
int v1 = it1.Peek().data,
v2 = it2.Peek().data;
// Base case
if (v1 + v2 == x)
return true;
// Moving forward pointer
if (v1 + v2 < x)
{
c = it1.Peek().right;
it1.Pop();
while (c != null)
{
it1.Push(c);
c = c.left;
}
}
// Moving backward pointer
else
{
c = it2.Peek().left;
it2.Pop();
while (c != null)
{
it2.Push(c);
c = c.right;
}
}
}
// Case when no pair is found
return false;
}
// Driver code
public static void Main(String[] args)
{
node root = new node(5);
root.left = new node(3);
root.right = new node(7);
root.left.left = new node(2);
root.left.right = new node(4);
root.right.left = new node(6);
root.right.right = new node(8);
int x = 5;
// Calling required function
if (existsPair(root, x))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by Rajput-Ji 输出：
Yes
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