Total sum except adjacent of a given node in BST

Given a BST and a key Node, find the total sum in BST, except those Node which are adjacent to key Node.
Examples:

1:-First find the total sum of BST
2:-Search the key Node and trace its parent Node.
3:-If the key Node is present then, subtract the sum of its adjacent Node from total sum
4:-If key is not present in BST then return -1.

C++

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// C++ program to find total sum except a given Node in BST
#include <bits/stdc++.h>
using namespace std;
  
struct Node {
    int data;
    struct Node *left, *right;
};
  
// insertion of Node in Tree
Node* getNode(int n)
{
    struct Node* root = new Node;
    root->data = n;
    root->left = NULL;
    root->right = NULL;
    return root;
}
  
// total sum of bst
int sum(struct Node* root)
{
    if (root == NULL)
        return 0;
  
    return root->data + sum(root->left) + sum(root->right);
}
  
// sum of all element except those which are adjecent to key Node
int adjSum(Node* root, int key)
{
  
    int parent = root->data;
  
    while (root != NULL) {
        if (key < root->data) {
            parent = root->data; 
            root = root->left;
        }
        else if (root->data == key) // key Node matches
        {
            // if the left Node and right Node of key is 
            // not null then add all adjecent Node and
            // substract from totalSum
            if (root->left != NULL && root->right != NULL)
                return (parent + root->left->data + 
                                 root->right->data);
  
            // if key is leaf
            if (root->left == NULL && root->right == NULL)
                return parent;
  
            // If only left child is null
            if (root->left == NULL) 
                return (parent + root->right->data);
  
            // If only right child is NULL
            if (root->right == NULL) 
                return (parent + root->left->data);
        }
  
        else {
            parent = root->data;
            root = root->right;
        }
    }
  
    return 0;
}
  
int findTotalExceptKey(Node *root, int key)
{
    return sum(root) - adjSum(root, key);
}
  
// Driver code
int main()
{
    struct Node* root = getNode(15);
    root->left = getNode(13);
    root->left->left = getNode(12);
    root->left->left->left = getNode(11);
    root->left->right = getNode(14);
    root->right = getNode(20);
    root->right->left = getNode(18);
    root->right->right = getNode(24);
    root->right->right->left = getNode(23);
    root->right->right->right = getNode(25);
    int key = 20;
    printf("%d ", findTotalExceptKey(root, key));
    return 0;
}

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Java

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// Java program to find total sum
// except a given Node in BST
class GFG
{
static class Node
{
    int data;
    Node left, right;
};
  
// insertion of Node in Tree
static Node getNode(int n)
{
    Node root = new Node();
    root.data = n;
    root.left = null;
    root.right = null;
    return root;
}
  
// total sum of bst
static int sum(Node root)
{
    if (root == null)
        return 0;
  
    return root.data + sum(root.left) + 
                       sum(root.right);
}
  
// sum of all element except those
// which are adjecent to key Node
static int adjSum(Node root, int key)
{
    int parent = root.data;
  
    while (root != null
    {
        if (key < root.data) 
        {
            parent = root.data; 
            root = root.left;
        }
        else if (root.data == key) // key Node matches
        {
            // if the left Node and right Node of key is 
            // not null then add all adjecent Node and
            // substract from totalSum
            if (root.left != null && root.right != null)
                return (parent + root.left.data + 
                                root.right.data);
  
            // if key is leaf
            if (root.left == null && root.right == null)
                return parent;
  
            // If only left child is null
            if (root.left == null
                return (parent + root.right.data);
  
            // If only right child is null
            if (root.right == null
                return (parent + root.left.data);
        }
        else 
        {
            parent = root.data;
            root = root.right;
        }
    }
    return 0;
}
  
static int findTotalExceptKey(Node root, int key)
{
    return sum(root) - adjSum(root, key);
}
  
// Driver code
public static void main(String[] args)
{
    Node root = getNode(15);
    root.left = getNode(13);
    root.left.left = getNode(12);
    root.left.left.left = getNode(11);
    root.left.right = getNode(14);
    root.right = getNode(20);
    root.right.left = getNode(18);
    root.right.right = getNode(24);
    root.right.right.left = getNode(23);
    root.right.right.right = getNode(25);
    int key = 20;
    System.out.printf("%d "
               findTotalExceptKey(root, key));
}
}
  
// This code is contributed by 29AjayKumar

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C#

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// C# program to find total sum
// except a given Node in BST
using System;
  
class GFG
{
class Node
{
    public int data;
    public Node left, right;
};
  
// insertion of Node in Tree
static Node getNode(int n)
{
    Node root = new Node();
    root.data = n;
    root.left = null;
    root.right = null;
    return root;
}
  
// total sum of bst
static int sum(Node root)
{
    if (root == null)
        return 0;
  
    return root.data + sum(root.left) + 
                       sum(root.right);
}
  
// sum of all element except those
// which are adjecent to key Node
static int adjSum(Node root, int key)
{
    int parent = root.data;
  
    while (root != null
    {
        if (key < root.data) 
        {
            parent = root.data; 
            root = root.left;
        }
        else if (root.data == key) // key Node matches
        {
            // if the left Node and right Node of key is 
            // not null then add all adjecent Node and
            // substract from totalSum
            if (root.left != null && root.right != null)
                return (parent + root.left.data + 
                                 root.right.data);
  
            // if key is leaf
            if (root.left == null && root.right == null)
                return parent;
  
            // If only left child is null
            if (root.left == null
                return (parent + root.right.data);
  
            // If only right child is null
            if (root.right == null
                return (parent + root.left.data);
        }
        else
        {
            parent = root.data;
            root = root.right;
        }
    }
    return 0;
}
  
static int findTotalExceptKey(Node root, int key)
{
    return sum(root) - adjSum(root, key);
}
  
// Driver code
public static void Main(String[] args)
{
    Node root = getNode(15);
    root.left = getNode(13);
    root.left.left = getNode(12);
    root.left.left.left = getNode(11);
    root.left.right = getNode(14);
    root.right = getNode(20);
    root.right.left = getNode(18);
    root.right.right = getNode(24);
    root.right.right.left = getNode(23);
    root.right.right.right = getNode(25);
    int key = 20;
    Console.Write("{0} "
            findTotalExceptKey(root, key));
}
}
  
// This code is contributed by PrinciRaj1992

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Output:

118

Time Complexity : O(n) + O(h) where n is number of nodes in BST and h is height of BST. We can write time complexity as O(n).

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Improved By : 29AjayKumar, princiraj1992

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