给定二叉搜索树和数字X。任务是找到BST中不同节点的不同对的数量，其总和等于X。没有两个节点具有相同的值。
例子：
Input : X = 5
5
/ \
3 7
/ \ / \
2 4 6 8
Output : 1
{2, 3} is the only possible pair.
Thus, the answer is equal to 1.
Input : X = 6
1
\
2
\
3
\
4
\
5
Output : 2
Possible pairs are {{1, 5}, {2, 4}}.
天真的方法：想法是散列BST的所有元素或将BST转换为排序的数组。之后，使用此处给出的算法找到对的数量。
时间复杂度： O(N)。
空间复杂度： O(N)。
空间优化方法：想法是在BST上使用两个指针技术。维护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。
- 为BST创建前进和后退迭代器。假设它们指向的节点的值分别等于v 1和v 2 。
- 现在在每一步
- 如果v 1 + v 2 = X，则找到该对，因此将计数增加1。
- 如果v 1 + v 2小于或等于x，我们将使迭代器指向下一个元素。
- 如果v 1 + v 2大于x，我们将使反向迭代器指向上一个元素。
- 重复上述步骤，直到两个迭代器都指向同一节点为止。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Node of 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
int cntPairs(node* root, int x)
{
// Stack to store nodes for
// forward and backward iterator
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;
// Variable to store final answer
int ans = 0;
// two pointer technique
while (it1.top() != it2.top()) {
// Variables to store the
// value of the nodes current
// iterators are pointing to.
int v1 = it1.top()->data;
int v2 = it2.top()->data;
// If we find a pair
// then count is increased by 1
if (v1 + v2 == x)
ans++;
// 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;
}
}
// Returning final answer
return ans;
}
// Driver code
int main()
{
/* 5
/ \
3 7
/ \ / \
2 4 6 8 */
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 = 10;
cout << cntPairs(root, x);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Node of 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
static int cntPairs(node root, int x)
{
// Stack to store nodes for
// forward and backward iterator
Stack it1 = new Stack<>();
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;
}
// Variable to store final answer
int ans = 0;
// two pointer technique
while (it1.peek() != it2.peek())
{
// Variables to store the
// value of the nodes current
// iterators are pointing to.
int v1 = it1.peek().data;
int v2 = it2.peek().data;
// If we find a pair
// then count is increased by 1
if (v1 + v2 == x)
ans++;
// 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;
}
}
}
// Returning final answer
return ans;
}
// Driver code
public static void main(String[] args)
{
/* 5
/ \
3 7
/ \ / \
2 4 6 8 */
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 = 10;
System.out.print(cntPairs(root, x));
}
}
// This code is contributed by Rajput-Ji Python
# Python implementation of above algorithm
# Utility class to create a node
class node:
def __init__(self, key):
self.data = key
self.left = self.right = None
# Function to find a pair
def cntPairs( root, x):
# Stack to store nodes for
# forward and backward iterator
it1 = []
it2 = []
# Initializing forward iterator
c = root
while (c != None):
it1.append(c)
c = c.left
# Initializing backward iterator
c = root
while (c != None):
it2.append(c)
c = c.right
# Variable to store final answer
ans = 0
# two pointer technique
while (it1[-1] != it2[-1]) :
# Variables to store the
# value of the nodes current
# iterators are pointing to.
v1 = it1[-1].data
v2 = it2[-1].data
# If we find a pair
# then count is increased by 1
if (v1 + v2 == x):
ans=ans+1
# Moving forward iterator
if (v1 + v2 <= x) :
c = it1[-1].right
it1.pop()
while (c != None):
it1.append(c)
c = c.left
# Moving backward iterator
else :
c = it2[-1].left
it2.pop()
while (c != None):
it2.append(c)
c = c.right
# Returning final answer
return ans
# Driver code
# 5
# / \
# 3 7
# / \ / \
# 2 4 6 8
root = node(5)
root.left = node(3)
root.right = node(7)
root.left.left = node(2)
root.left.right = node(4)
root.right.left = node(6)
root.right.right = node(8)
x = 10
print(cntPairs(root, x))
# This code is contributed by Arnab KunduC#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Node of 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
static int cntPairs(node root, int x)
{
// Stack to store nodes for
// forward and backward iterator
Stack it1 = new Stack();
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;
}
// Variable to store readonly answer
int ans = 0;
// two pointer technique
while (it1.Peek() != it2.Peek())
{
// Variables to store the
// value of the nodes current
// iterators are pointing to.
int v1 = it1.Peek().data;
int v2 = it2.Peek().data;
// If we find a pair
// then count is increased by 1
if (v1 + v2 == x)
ans++;
// 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;
}
}
}
// Returning readonly answer
return ans;
}
// Driver code
public static void Main(String[] args)
{
/* 5
/ \
3 7
/ \ / \
2 4 6 8 */
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 = 10;
Console.Write(cntPairs(root, x));
}
}
// This code is contributed by Rajput-Ji 输出：
3
时间复杂度： O(N)
空间复杂度： O(H)，其中H是BST的高度
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