Sum of all nodes with smaller values at a distance K from a given node in a BST

Given a Binary Search Tree, a target node in the BST, and an integer value K, the task is to find the sum of all nodes that are at a distance K from the target node whose value is less than the target node.

Examples:

Input: target = 7, K = 2

Output: 11
Explanation:
The nodes at a distance K(= 2) from the node 7 is 1, 4, and 6. Therefore, the sum of nodes is 11.

Input: target = 5, K = 1

Output: 4

Approach: The given problem can be solved by performing DFS Traversal for K distance below the target node and perform the DFS Traversal upward K distance from the target node. Follow the steps below to solve the problem:

Define a function kDistanceDownSum(root, k, &sum) and perform the following steps:
- For the Base Case, check if the root is nullptr and k is less than 0, then return from the function.
- If the value of k equals 0, then add root->val to the variable sum and return.
- Call the same function kDistanceDownSum(root->left, k-1, sum) and kDistanceDownSum(root->right, k – 1, sum) for the left and right sub-trees.
For the Base Case, check if the root is nullptr, then return -1.
If the root is the same as the target, then call the function kDistanceDownSum(root->left, k – 1, sum) to calculate the sum for the first type of nodes and return 0(No second type of nodes possible).
Initialize the variable dl as -1 and if the target is less than root, then set the value of dl as the value returned by the function kDistanceSum(root->left, target k, sum).
If the value of dl is not equal to -1, then if sum equals (dl + 1), then add the value of root->data to the sum and then return -1.
Similarly, initialize the variable dr as -1 and if the target is greater than the root, then update the value of dr to the value returned by kDistanceSum(root->right, target k, sum).
If the value of dr is not equal to -1, then if the value of sum equals (dr + 1), then add the value of root->data to the sum. Otherwise, call the function kDistanceDownSum(root->left, k – dr – 2, sum) and return (1 + dr).
After performing the above steps, print the value of ans as the resultant sum.

Following is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Structure of Tree
struct TreeNode {
 
    int data;
    TreeNode* left;
    TreeNode* right;
 
    // Constructor
    TreeNode(int data)
    {
        this->data = data;
        this->left = NULL;
        this->right = NULL;
    }
};
 
// Function to add the node to the sum
// below the target node
void kDistanceDownSum(TreeNode* root,
                      int k, int& sum)
{
 
    // Base Case
    if (root == NULL || k < 0)
        return;
 
    // If Kth distant node is reached
    if (k == 0) {
        sum += root->data;
        return;
    }
 
    // Recur for the left and the
    // right subtrees
    kDistanceDownSum(root->left,
                     k - 1, sum);
    kDistanceDownSum(root->right,
                     k - 1, sum);
}
 
// Function to find the K distant nodes
// from target node, it returns -1 if
// target node is not present in tree
int kDistanceSum(TreeNode* root,
                 int target,
                 int k, int& sum)
{
    // Base Case 1
    if (root == NULL)
        return -1;
 
    // If target is same as root.
    if (root->data == target) {
        kDistanceDownSum(root->left,
                         k - 1, sum);
        return 0;
    }
 
    // Recur for the left subtree
    int dl = -1;
 
    // Tree is BST so reduce the
    // search space
    if (target < root->data) {
        dl = kDistanceSum(root->left,
                          target, k, sum);
    }
 
    // Check if target node was found
    // in left subtree
    if (dl != -1) {
 
        // If root is at distance k from
        // the target
        if (dl + 1 == k)
            sum += root->data;
 
        // Node less than target will be
        // present in left
        return -1;
    }
 
    // When node is not present in the
    // left subtree
    int dr = -1;
    if (target > root->data) {
        dr = kDistanceSum(root->right,
                          target, k, sum);
    }
 
    if (dr != -1) {
 
        // If Kth distant node is reached
        if (dr + 1 == k)
            sum += root->data;
 
        // Node less than target at k
        // distance maybe present in the
        // left tree
        else
            kDistanceDownSum(root->left,
                             k - dr - 2, sum);
 
        return 1 + dr;
    }
 
    // If target was not present in the
    // left nor in right subtree
    return -1;
}
 
// Function to insert a node in BST
TreeNode* insertNode(int data,
                     TreeNode* root)
{
    // If root is NULL
    if (root == NULL) {
        TreeNode* node = new TreeNode(data);
        return node;
    }
 
    // Insert the data in right half
    else if (data > root->data) {
        root->right = insertNode(
            data, root->right);
    }
 
    // Insert the data in left half
    else if (data <= root->data) {
        root->left = insertNode(
            data, root->left);
    }
 
    // Return the root node
    return root;
}
 
// Function to find the sum of K distant
// nodes from the target node having
// value less than target node
void findSum(TreeNode* root, int target,
             int K)
{
 
    // Stores the sum of nodes having
    // values < target at K distance
    int sum = 0;
 
    kDistanceSum(root, target, K, sum);
 
    // Print the resultant sum
    cout << sum;
}
 
// Driver Code
int main()
{
    TreeNode* root = NULL;
    int N = 11;
    int tree[] = { 3, 1, 7, 0, 2, 5,
                   10, 4, 6, 9, 8 };
 
    // Create the Tree
    for (int i = 0; i < N; i++) {
        root = insertNode(tree[i], root);
    }
 
    int target = 7;
    int K = 2;
    findSum(root, target, K);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
public class GFG{
    static int sum;
   
// Structure of Tree
static class TreeNode {
 
    int data;
    TreeNode left;
    TreeNode right;
 
    // Constructor
    TreeNode(int data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
};
 
// Function to add the node to the sum
// below the target node
static void kDistanceDownSum(TreeNode root,
                      int k)
{
 
    // Base Case
    if (root == null || k < 0)
        return;
 
    // If Kth distant node is reached
    if (k == 0) {
        sum += root.data;
        return;
    }
 
    // Recur for the left and the
    // right subtrees
    kDistanceDownSum(root.left,
                     k - 1);
    kDistanceDownSum(root.right,
                     k - 1);
}
 
// Function to find the K distant nodes
// from target node, it returns -1 if
// target node is not present in tree
static int kDistanceSum(TreeNode root,
                 int target,
                 int k)
{
    // Base Case 1
    if (root == null)
        return -1;
 
    // If target is same as root.
    if (root.data == target) {
        kDistanceDownSum(root.left,
                         k - 1);
    return 0;
    }
 
    // Recur for the left subtree
    int dl = -1;
 
    // Tree is BST so reduce the
    // search space
    if (target < root.data) {
        dl = kDistanceSum(root.left,
                          target, k);
    }
 
    // Check if target node was found
    // in left subtree
    if (dl != -1) {
 
        // If root is at distance k from
        // the target
        if (dl + 1 == k)
            sum += root.data;
 
        // Node less than target will be
        // present in left
        return -1;
    }
 
    // When node is not present in the
    // left subtree
    int dr = -1;
    if (target > root.data) {
        dr = kDistanceSum(root.right,
                          target, k);
    }
 
    if (dr != -1) {
 
        // If Kth distant node is reached
        if (dr + 1 == k)
            sum += root.data;
 
        // Node less than target at k
        // distance maybe present in the
        // left tree
        else
            kDistanceDownSum(root.left,
                             k - dr - 2);
 
        return 1 + dr;
    }
 
    // If target was not present in the
    // left nor in right subtree
    return -1;
}
 
// Function to insert a node in BST
static TreeNode insertNode(int data,
                     TreeNode root)
{
    // If root is null
    if (root == null) {
        TreeNode node = new TreeNode(data);
        return node;
    }
 
    // Insert the data in right half
    else if (data > root.data) {
        root.right = insertNode(
            data, root.right);
    }
 
    // Insert the data in left half
    else if (data <= root.data) {
        root.left = insertNode(
            data, root.left);
    }
 
    // Return the root node
    return root;
}
 
// Function to find the sum of K distant
// nodes from the target node having
// value less than target node
static void findSum(TreeNode root, int target,
             int K)
{
 
    // Stores the sum of nodes having
    // values < target at K distance
    sum = 0;
 
    kDistanceSum(root, target, K);
 
    // Print the resultant sum
    System.out.print(sum);
}
 
// Driver Code
public static void main(String[] args)
{
    TreeNode root = null;
    int N = 11;
    int tree[] = { 3, 1, 7, 0, 2, 5,
                   10, 4, 6, 9, 8 };
 
    // Create the Tree
    for (int i = 0; i < N; i++) {
        root = insertNode(tree[i], root);
    }
 
    int target = 7;
    int K = 2;
    findSum(root, target, K);
 
}
}
 
// This code is contributed by 29AjayKumar 

Python3

# python 3 program for the above approach
 
# Structure of Tree
sum = 0
 
class Node:
    # A constructor to create a new node
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
 
# Function to add the node to the sum
# below the target node
def kDistanceDownSum(root, k):
    global sum
    # Base Case
    if (root == None or k < 0):
        return
 
    # If Kth distant node is reached
    if (k == 0):
        sum += root.data
        return
 
    # Recur for the left and the
    # right subtrees
    kDistanceDownSum(root.left,k - 1)
    kDistanceDownSum(root.right,k - 1)
 
# Function to find the K distant nodes
# from target node, it returns -1 if
# target node is not present in tree
def kDistanceSum(root, target, k):
    global sum
    # Base Case 1
    if (root == None):
        return -1
 
    # If target is same as root.
    if (root.data == target):
        kDistanceDownSum(root.left,k - 1)
        return 0
 
    # Recur for the left subtree
    dl = -1
 
    # Tree is BST so reduce the
    # search space
    if (target < root.data):
        dl = kDistanceSum(root.left, target, k)
 
    # Check if target node was found
    # in left subtree
    if (dl != -1):
        # If root is at distance k from
        # the target
        if (dl + 1 == k):
            sum += root.data
 
        # Node less than target will be
        # present in left
        return -1
 
    # When node is not present in the
    # left subtree
    dr = -1
    if (target > root.data):
        dr = kDistanceSum(root.right, target, k)
 
    if (dr != -1):
        # If Kth distant node is reached
        if (dr + 1 == k):
            sum += root.data
 
        # Node less than target at k
        # distance maybe present in the
        # left tree
        else:
            kDistanceDownSum(root.left, k - dr - 2)
 
        return 1 + dr
 
    # If target was not present in the
    # left nor in right subtree
    return -1
 
# Function to insert a node in BST
def insertNode(data, root):
    # If root is NULL
    if (root == None):
        node = Node(data)
        return node
 
    # Insert the data in right half
    elif (data > root.data):
        root.right = insertNode(data, root.right)
 
    # Insert the data in left half
    elif(data <= root.data):
        root.left = insertNode(data, root.left)
 
    # Return the root node
    return root
 
# Function to find the sum of K distant
# nodes from the target node having
# value less than target node
def findSum(root, target, K):
   
    # Stores the sum of nodes having
    # values < target at K distance
    kDistanceSum(root, target, K)
 
    # Print the resultant sum
    print(sum)
 
# Driver Code
if __name__ == '__main__':
    root = None
    N = 11
    tree = [3, 1, 7, 0, 2, 5,10, 4, 6, 9, 8]
 
    # Create the Tree
    for i in range(N):
        root = insertNode(tree[i], root)
 
    target = 7
    K = 2
    findSum(root, target, K)
     
    # This code is contributed by SURENDRA_GANGWAR.

C#

// C# program for the above approach
using System;
 
public class GFG{
    static int sum;
   
// Structure of Tree
public
 
 class TreeNode {
 
    public
 
 int data;
    public
 
 TreeNode left;
    public
 
 TreeNode right;
 
    // Constructor
    public TreeNode(int data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
};
 
// Function to add the node to the sum
// below the target node
static void kDistanceDownSum(TreeNode root,
                      int k)
{
 
    // Base Case
    if (root == null || k < 0)
        return;
 
    // If Kth distant node is reached
    if (k == 0) {
        sum += root.data;
        return;
    }
 
    // Recur for the left and the
    // right subtrees
    kDistanceDownSum(root.left,
                     k - 1);
    kDistanceDownSum(root.right,
                     k - 1);
}
 
// Function to find the K distant nodes
// from target node, it returns -1 if
// target node is not present in tree
static int kDistanceSum(TreeNode root,
                 int target,
                 int k)
{
    // Base Case 1
    if (root == null)
        return -1;
 
    // If target is same as root.
    if (root.data == target) {
        kDistanceDownSum(root.left,
                         k - 1);
    return 0;
    }
 
    // Recur for the left subtree
    int dl = -1;
 
    // Tree is BST so reduce the
    // search space
    if (target < root.data) {
        dl = kDistanceSum(root.left,
                          target, k);
    }
 
    // Check if target node was found
    // in left subtree
    if (dl != -1) {
 
        // If root is at distance k from
        // the target
        if (dl + 1 == k)
            sum += root.data;
 
        // Node less than target will be
        // present in left
        return -1;
    }
 
    // When node is not present in the
    // left subtree
    int dr = -1;
    if (target > root.data) {
        dr = kDistanceSum(root.right,
                          target, k);
    }
 
    if (dr != -1) {
 
        // If Kth distant node is reached
        if (dr + 1 == k)
            sum += root.data;
 
        // Node less than target at k
        // distance maybe present in the
        // left tree
        else
            kDistanceDownSum(root.left,
                             k - dr - 2);
 
        return 1 + dr;
    }
 
    // If target was not present in the
    // left nor in right subtree
    return -1;
}
 
// Function to insert a node in BST
static TreeNode insertNode(int data,
                     TreeNode root)
{
    // If root is null
    if (root == null) {
        TreeNode node = new TreeNode(data);
        return node;
    }
 
    // Insert the data in right half
    else if (data > root.data) {
        root.right = insertNode(
            data, root.right);
    }
 
    // Insert the data in left half
    else if (data <= root.data) {
        root.left = insertNode(
            data, root.left);
    }
 
    // Return the root node
    return root;
}
 
// Function to find the sum of K distant
// nodes from the target node having
// value less than target node
static void findSum(TreeNode root, int target,
             int K)
{
 
    // Stores the sum of nodes having
    // values < target at K distance
    sum = 0;
 
    kDistanceSum(root, target, K);
 
    // Print the resultant sum
    Console.Write(sum);
}
 
// Driver Code
public static void Main(String[] args)
{
    TreeNode root = null;
    int N = 11;
    int []tree = { 3, 1, 7, 0, 2, 5,
                   10, 4, 6, 9, 8 };
 
    // Create the Tree
    for (int i = 0; i < N; i++) {
        root = insertNode(tree[i], root);
    }
 
    int target = 7;
    int K = 2;
    findSum(root, target, K);
 
}
}
 
// This code is contributed by gauravrajput1 

Javascript

<script>
// Javascript program for the above approach
 
// Structure of Tree
let sum = 0;
 
class TreeNode {
  // Constructor
  constructor(data = "", left = null, right = null) {
    this.data = data;
    this.left = left;
    this.right = right;
  }
}
 
// Function to add the node to the sum
// below the target node
function kDistanceDownSum(root, k)
{
  // Base Case
  if (root == null || k < 0) {
    return
  }
 
  // If Kth distant node is reached
  if (k == 0) {
    sum += root.data;
    return;
  }
 
  // Recur for the left and the
  // right subtrees
  kDistanceDownSum(root.left, k - 1);
  kDistanceDownSum(root.right, k - 1);
}
 
// Function to find the K distant nodes
// from target node, it returns -1 if
// target node is not present in tree
function kDistanceSum(root, target, k) {
  // Base Case 1
  if (root == null) return -1;
 
  // If target is same as root.
  if (root.data == target) {
    kDistanceDownSum(root.left, k - 1);
    return 0;
  }
 
  // Recur for the left subtree
  let dl = -1;
 
  // Tree is BST so reduce the
  // search space
  if (target < root.data) {
    dl = kDistanceSum(root.left, target, k);
  }
 
  // Check if target node was found
  // in left subtree
  if (dl != -1) {
    // If root is at distance k from
    // the target
    if (dl + 1 == k) sum += root.data;
 
    // Node less than target will be
    // present in left
    return -1;
  }
 
  // When node is not present in the
  // left subtree
  let dr = -1;
  if (target > root.data) {
    dr = kDistanceSum(root.right, target, k);
  }
 
  if (dr != -1) {
    // If Kth distant node is reached
    if (dr + 1 == k) sum += root.data;
    // Node less than target at k
    // distance maybe present in the
    // left tree
    else kDistanceDownSum(root.left, k - dr - 2);
 
    return 1 + dr;
  }
 
  // If target was not present in the
  // left nor in right subtree
  return -1;
}
 
// Function to insert a node in BST
function insertNode(data, root) {
  // If root is null
  if (root == null) {
    let node = new TreeNode(data);
    return node;
  }
 
  // Insert the data in right half
  else if (data > root.data) {
    root.right = insertNode(data, root.right);
  }
 
  // Insert the data in left half
  else if (data <= root.data) {
    root.left = insertNode(data, root.left);
  }
 
  // Return the root node
  return root;
}
 
// Function to find the sum of K distant
// nodes from the target node having
// value less than target node
function findSum(root, target, K) {
  // Stores the sum of nodes having
  // values < target at K distance
  kDistanceSum(root, target, K, sum);
 
  // Print the resultant sum
  document.write(sum);
}
 
// Driver Code
 
let root = null;
let N = 11;
let tree = [3, 1, 7, 0, 2, 5, 10, 4, 6, 9, 8];
 
// Create the Tree
for (let i = 0; i < N; i++) {
  root = insertNode(tree[i], root);
}
 
let target = 7;
let K = 2;
findSum(root, target, K);
</script>

Output

Time Complexity: O(N) where N is the number of nodes in given binary tree.
Auxiliary Space: O(h) where h is the height of binary tree due to recursion call stack.

Approach using BFS:-

We will be using level order traversal to find the sum of smaller value nodes

Implementation:-

First we will find the target node using level order traversal.

While finding the target node we will store the parent of each node so that we can move towards the parent of the node as well.

After this we will traverse from the target node to all the tree directions that is toward both child and parent till distance K and add the values of node into our answer which are smaller than target at distance K.

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Structure of Tree
struct TreeNode {
 
    int data;
    TreeNode* left;
    TreeNode* right;
 
    // Constructor
    TreeNode(int data)
    {
        this->data = data;
        this->left = NULL;
        this->right = NULL;
    }
};
 
// Function to insert a node in BST
TreeNode* insertNode(int data,
                     TreeNode* root)
{
    // If root is NULL
    if (root == NULL) {
        TreeNode* node = new TreeNode(data);
        return node;
    }
 
    // Insert the data in right half
    else if (data > root->data) {
        root->right = insertNode(
            data, root->right);
    }
 
    // Insert the data in left half
    else if (data <= root->data) {
        root->left = insertNode(
            data, root->left);
    }
 
    // Return the root node
    return root;
}
 
// Function to find the sum of K distant
// nodes from the target node having
// value less than target node
void findSum(TreeNode* root, int target,
             int K)
{
    //variable to store answer
    int ans = 0;
 
    //queue for bfs
    queue<TreeNode*> q;
 
    q.push(root);
 
    //to store target node
    TreeNode* need;
 
    //map to store parent of each node
    unordered_map<TreeNode*, TreeNode*> m;
 
    //bfs
    while(q.size()){
 
      int s = q.size();
 
      //traversing to current level
      for(int i=0;i<s;i++){
 
        TreeNode* temp = q.front();
 
        q.pop();
 
        //if target value found
        if(temp->data==target) need=temp;
 
        if(temp->left){
          q.push(temp->left);
          m[temp->left]=temp;
        }
 
        if(temp->right){
          q.push(temp->right);
          m[temp->right]=temp;
        }
 
      }
 
    }
 
    //map to store occurrence of a node
    //that is the node has taken or not
    unordered_map<TreeNode*, int> mm;
 
    q.push(need);
 
    //to store current distance
    int c = 0;
 
    while(q.size()){
 
      int s = q.size();
 
      for(int i=0;i<s;i++){
 
        TreeNode* temp = q.front();
 
        q.pop();
 
        mm[temp] = 1;
         
        if(temp->data<target and c==K)
        ans+=temp->data;
 
        //moving left
        if(temp->left&&mm[temp->left]==0){
          q.push(temp->left);
        }
 
        //moving right
        if(temp->right&&mm[temp->right]==0){
          q.push(temp->right);
        }
 
        //movinf to parent
        if(m[temp]&&mm[m[temp]]==0){
          q.push(m[temp]);
        }
 
      }
 
      c++;
      if(c>K)break;
 
    }
    cout<<ans<<endl;
 
}
 
// Driver Code
int main()
{
    TreeNode* root = NULL;
    int N = 11;
    int tree[] = { 3, 1, 7, 0, 2, 5,
                   10, 4, 6, 9, 8 };
 
    // Create the Tree
    for (int i = 0; i < N; i++) {
        root = insertNode(tree[i], root);
    }
 
    int target = 7;
    int K = 2;
    findSum(root, target, K);
 
    return 0;
}
//code contributed by shubhamrajput6156

Java

import java.util.*;
 
// Structure of Tree
class TreeNode {
 
    int data;
    TreeNode left;
    TreeNode right;
 
    // Constructor
    TreeNode(int data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
 
public class Main {
 
    // Function to insert a node in BST
    public static TreeNode insertNode(int data,
                                      TreeNode root)
    {
        // If root is null
        if (root == null) {
            TreeNode node = new TreeNode(data);
            return node;
        }
 
        // Insert the data in right half
        else if (data > root.data) {
            root.right = insertNode(
                data, root.right);
        }
 
        // Insert the data in left half
        else if (data <= root.data) {
            root.left = insertNode(
                data, root.left);
        }
 
        // Return the root node
        return root;
    }
 
    // Function to find the sum of K distant
    // nodes from the target node having
    // value less than target node
    public static void findSum(TreeNode root, int target,
                               int K)
    {
        // variable to store answer
        int ans = 0;
 
        // queue for bfs
        Queue<TreeNode> q = new LinkedList<>();
 
        q.add(root);
 
        // to store target node
        TreeNode need = null;
 
        // map to store parent of each node
        HashMap<TreeNode, TreeNode> m = new HashMap<>();
 
        // bfs
        while (!q.isEmpty()) {
 
            int s = q.size();
 
            // traversing to current level
            for (int i = 0; i < s; i++) {
 
                TreeNode temp = q.peek();
                q.remove();
 
                // if target value found
                if (temp.data == target)
                    need = temp;
 
                if (temp.left != null) {
                    q.add(temp.left);
                    m.put(temp.left, temp);
                }
 
                if (temp.right != null) {
                    q.add(temp.right);
                    m.put(temp.right, temp);
                }
            }
        }
 
        // map to store occurrence of a node
        // that is the node has taken or not
        HashMap<TreeNode, Integer> mm = new HashMap<>();
 
        q.add(need);
 
        // to store current distance
        int c = 0;
 
        while (!q.isEmpty()) {
 
            int s = q.size();
 
            for (int i = 0; i < s; i++) {
 
                TreeNode temp = q.peek();
                q.remove();
 
                mm.put(temp, 1);
 
                if (temp.data < target && c == K)
                    ans += temp.data;
 
                // moving left
                if (temp.left != null && mm.getOrDefault(temp.left, 0) == 0) {
                    q.add(temp.left);
                }
 
                // moving right
                if (temp.right != null && mm.getOrDefault(temp.right, 0) == 0) {
                    q.add(temp.right);
                }
 
                // moving to parent
                if (m.get(temp) != null && mm.getOrDefault(m.get(temp), 0) == 0) {
                    q.add(m.get(temp));
                }
            }
 
            c++;
            if (c > K)
                break;
        }
        System.out.println(ans);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        TreeNode root = null;
        int N = 11;
        int[] tree = { 3, 1, 7, 0, 2, 5, 10, 4, 6, 9, 8 };
 
        // Create the Tree
        for (int i = 0; i < N; i++) {
            root = insertNode(tree[i], root);
        }
 
        int target = 7;
        int K = 2;
        findSum(root, target, K);
    }
}

Python

from collections import deque
 
# Structure of Tree
class TreeNode:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
 
# Function to insert a node in BST
def insertNode(data, root):
    # If root is None
    if root is None:
        return TreeNode(data)
 
    # Insert the data in the right half
    if data > root.data:
        root.right = insertNode(data, root.right)
 
    # Insert the data in the left half
    else:
        root.left = insertNode(data, root.left)
 
    # Return the root node
    return root
 
# Function to find the sum of K distant nodes from the target node having value less than target node
def findSum(root, target, K):
    # Variable to store the answer
    ans = 0
 
    # Queue for BFS
    q = deque()
    q.append(root)
 
    # To store the target node
    need = None
 
    # Dictionary to store parent of each node
    m = {}
 
    # BFS
    while q:
        s = len(q)
 
        # Traverse the current level
        for i in range(s):
            temp = q.popleft()
 
            # If the target value is found
            if temp.data == target:
                need = temp
 
            if temp.left:
                q.append(temp.left)
                m[temp.left] = temp
 
            if temp.right:
                q.append(temp.right)
                m[temp.right] = temp
 
    # Dictionary to store the occurrence of a node
    # that is whether the node has been visited or not
    mm = {}
    q.append(need)
 
    # To store the current distance
    c = 0
 
    while q:
        s = len(q)
 
        for i in range(s):
            temp = q.popleft()
            mm[temp] = 1
 
            if temp.data < target and c == K:
                ans += temp.data
 
            # Moving left
            if temp.left and mm.get(temp.left, 0) == 0:
                q.append(temp.left)
 
            # Moving right
            if temp.right and mm.get(temp.right, 0) == 0:
                q.append(temp.right)
 
            # Moving to parent
            if temp in m and mm.get(m[temp], 0) == 0:
                q.append(m[temp])
 
        c += 1
        if c > K:
            break
 
    print(ans)
 
# Driver Code
if __name__ == "__main__":
    root = None
    N = 11
    tree = [3, 1, 7, 0, 2, 5, 10, 4, 6, 9, 8]
 
    # Create the Tree
    for i in range(N):
        root = insertNode(tree[i], root)
 
    target = 7
    K = 2
    findSum(root, target, K)

C#

using System;
using System.Collections.Generic;
 
namespace BinaryTreeKDistanceSum
{
    class TreeNode
    {
        public int data;
        public TreeNode left;
        public TreeNode right;
 
        // Constructor
        public TreeNode(int data)
        {
            this.data = data;
            this.left = null;
            this.right = null;
        }
    }
 
    class Program
    {
        static TreeNode InsertNode(int data, TreeNode root)
        {
            if (root == null)
            {
                TreeNode node = new TreeNode(data);
                return node;
            }
            else if (data > root.data)
            {
                root.right = InsertNode(data, root.right);
            }
            else if (data <= root.data)
            {
                root.left = InsertNode(data, root.left);
            }
            return root;
        }
 
        static void FindSum(TreeNode root, int target, int K)
        {
            int ans = 0;
            Queue<TreeNode> q = new Queue<TreeNode>();
            q.Enqueue(root);
 
            TreeNode need = null;
            Dictionary<TreeNode, TreeNode> m = new Dictionary<TreeNode, TreeNode>();
            while (q.Count > 0)
            {
                int s = q.Count;
                for (int i = 0; i < s; i++)
                {
                    TreeNode temp = q.Dequeue();
 
                    if (temp.data == target) need = temp;
 
                    if (temp.left != null)
                    {
                        q.Enqueue(temp.left);
                        m[temp.left] = temp;
                    }
                    if (temp.right != null)
                    {
                        q.Enqueue(temp.right);
                        m[temp.right] = temp;
                    }
                }
            }
 
            Dictionary<TreeNode, int> mm = new Dictionary<TreeNode, int>();
            q.Enqueue(need);
            int c = 0;
            while (q.Count > 0)
            {
                int s = q.Count;
                for (int i = 0; i < s; i++)
                {
                    TreeNode temp = q.Dequeue();
                    mm[temp] = 1;
 
                    if (temp.data < target && c == K)
                    {
                        ans += temp.data;
                    }
 
                    if (temp.left != null && !mm.ContainsKey(temp.left))
                    {
                        q.Enqueue(temp.left);
                    }
                    if (temp.right != null && !mm.ContainsKey(temp.right))
                    {
                        q.Enqueue(temp.right);
                    }
                    if (m.ContainsKey(temp) && !mm.ContainsKey(m[temp]))
                    {
                        q.Enqueue(m[temp]);
                    }
                }
                c++;
                if (c > K) break;
            }
            Console.WriteLine(ans);
        }
 
        static void Main(string[] args)
        {
            TreeNode root = null;
            int N = 11;
            int[] tree = { 3, 1, 7, 0, 2, 5, 10, 4, 6, 9, 8 };
 
            for (int i = 0; i < N; i++)
            {
                root = InsertNode(tree[i], root);
            }
 
            int target = 7;
            int K = 2;
            FindSum(root, target, K);
        }
    }
}

Javascript

class TreeNode {
    constructor(data) {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
 
// Function to insert a node into the Binary Search Tree (BST)
function insertNode(data, root) {
    if (root === null) {
        const node = new TreeNode(data);
        return node;
    }
 
    if (data > root.data) {
        root.right = insertNode(data, root.right);
    } else if (data <= root.data) {
        root.left = insertNode(data, root.left);
    }
 
    return root;
}
 
// Function to find the sum of K distant nodes from the target
// node having values less than the target node
function findSum(root, target, K) {
    let ans = 0; // Variable to store the sum
    const queue = [root]; // Initialize a queue for BFS
    let need = null; // Node to store the target node
    const parentMap = new Map(); // Map to store the parent of each node
 
    // Perform BFS to find the target node and build the parent map
    while (queue.length > 0) {
        const levelSize = queue.length;
 
        for (let i = 0; i < levelSize; i++) {
            const temp = queue.shift();
 
            if (temp.data === target) {
                need = temp; // Found the target node
            }
 
            if (temp.left !== null) {
                queue.push(temp.left);
                parentMap.set(temp.left, temp);
            }
 
            if (temp.right !== null) {
                queue.push(temp.right);
                parentMap.set(temp.right, temp);
            }
        }
    }
 
    const mm = new Map(); // Map to track visited nodes
    queue.push(need); // Initialize the queue with the target node
    let distance = 0; // Initialize the distance from the target node
 
    // Perform BFS to calculate the sum of nodes at distance K from the target node
    while (queue.length > 0) {
        const levelSize = queue.length;
 
        for (let i = 0; i < levelSize; i++) {
            const temp = queue.shift();
            mm.set(temp, 1); // Mark the node as visited
 
            if (temp.data < target && distance === K) {
                ans += temp.data; // Add the node's value to the sum if it meets the criteria
            }
 
            if (temp.left !== null && mm.get(temp.left) !== 1) {
                queue.push(temp.left); // Explore the left child if not visited
            }
 
            if (temp.right !== null && mm.get(temp.right) !== 1) {
                queue.push(temp.right); // Explore the right child if not visited
            }
 
            if (parentMap.has(temp) && mm.get(parentMap.get(temp)) !== 1) {
                queue.push(parentMap.get(temp)); // Explore the parent if not visited
            }
        }
 
        distance++;
 
        if (distance > K) {
            break; // Stop the BFS when the desired distance is reached
        }
    }
 
    console.log(ans); // Print the final sum
}
 
// Driver Code
function main() {
    let root = null;
    const N = 11;
    const tree = [3, 1, 7, 0, 2, 5, 10, 4, 6, 9, 8];
 
    // Create the BST by inserting elements from the 'tree' array
    for (let i = 0; i < N; i++) {
        root = insertNode(tree[i], root);
    }
 
    const target = 7;
    const K = 2;
    findSum(root, target, K); // Find and display the sum
}
 
main();

Output

Time Complexity:- O(N)
Auxiliary Space:- O(N)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!

Sum of all nodes with smaller values at a distance K from a given node in a BST

C++

Java

Python3

C#

Javascript

C++

Java

Python

C#

Javascript

LangExtract + Milvus: A Practical Guide to Building a Hybrid Document Processing and Search System

Stop Your AI Assistant from Writing Outdated Code with Milvus SDK Code Helper

A Practical Guide for Choosing the Right Vector Database for Your AI Applications

LEAVE A REPLY Cancel reply

Most Popular

Why I’d skip the carrier store for this Google Store–only Pixel 10 setup perk

A big change is coming to OnePlus’s cameras

The Pixel 10 has serious pairing issues with the Samsung Galaxy Watch

Samsung Galaxy S26 Ultra could get bigger and rounder

EDITOR PICKS

Why I’d skip the carrier store for this Google Store–only Pixel 10 setup perk

A big change is coming to OnePlus’s cameras

The Pixel 10 has serious pairing issues with the Samsung Galaxy Watch

POPULAR POSTS

Why I’d skip the carrier store for this Google Store–only Pixel 10 setup perk

A big change is coming to OnePlus’s cameras

The Pixel 10 has serious pairing issues with the Samsung Galaxy Watch

POPULAR CATEGORY

ABOUT US

FOLLOW US