与 BST 中的给定总和配对 | 2套

给定一个二叉搜索树和一个整数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Yes否则打印No 。

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，我们将使用空间高效的方法通过将空间复杂度降低到 O(H) 来解决相同的问题，其中 H 是 BST 的高度。为此，我们将在 BST 上使用两个指针技术。因此，我们将维护一个前向和后向迭代器，它们将分别按照中序和反向中序遍历的顺序迭代 BST。以下是解决问题的步骤：

1. 为 BST 创建一个向前和向后迭代器。假设它们指向的节点的值是 v1 和 v2。
2. 现在每一步，
   • 如果 v1 + v2 = X，我们找到了一对。
   • 如果 v1 + v2 < x，我们将使前向迭代器指向下一个元素。
   • 如果 v1 + v2 > x，我们将使向后迭代器指向前一个元素。
3. 如果我们找不到这样的一对，答案将是“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 29

C#

// 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

Javascript

 
// Javascript implementation of the approach
 
// Node of the binary tree
class node
{
    constructor(data)
    {
        this.data = data;
        this.left = this.right = null;
    }
}
 
// Function to find a pair with given sum
function existsPair(root, x)
{
     
    // Iterators for BST
    let it1 = [], it2 = [];
  
    // Initializing forward iterator
    let 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[it1.length-1] != it2[it2.length-1])
    {
  
        // Variables to store values at
        // it1 and it2
        let v1 = it1[it1.length - 1].data,
            v2 = it2[it2.length - 1].data;
  
        // Base case
        if (v1 + v2 == x)
            return true;
  
        // Moving forward pointer
        if (v1 + v2 < x)
        {
            c = it1[it1.length - 1].right;
            it1.pop();
             
            while (c != null)
            {
                it1.push(c);
                c = c.left;
            }
        }
  
        // Moving backward pointer
        else
        {
            c = it2[it2.length - 1].left;
            it2.pop();
             
            while (c != null)
            {
                it2.push(c);
                c = c.right;
            }
        }
    }
      
    // Case when no pair is found
    return false;
}
 
// Driver code
let 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);
 
let x = 5;
 
// Calling required function
if (existsPair(root, x))
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by unknown2108
 

输出：

Yes

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