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K Dimensional Tree | Set 2 (Find Minimum)

  • Difficulty Level : Medium
  • Last Updated : 03 Nov, 2020

We strongly recommend to refer below post as a prerequisite of this.

K Dimensional Tree | Set 1 (Search and Insert)

In this post find minimum is discussed. The operation is to find minimum in the given dimension. This is especially needed in delete operation.

For example, consider below KD Tree, if given dimension is x, then output should be 5 and if given dimensions is y, then output should be 12.


In KD tree, points are divided dimension by dimension. For example, root divides keys by dimension 0, level next to root divides by dimension 1, next level by dimension 2 if k is more then 2 (else by dimension 0), and so on.

To find minimum we traverse nodes starting from root. If dimension of current level is same as given dimension, then required minimum lies on left side if there is left child. This is same as Binary Search Tree Minimum.
Above is simple, what to do when current level’s dimension is different. When dimension of current level is different, minimum may be either in left subtree or right subtree or current node may also be minimum. So we take minimum of three and return. This is different from Binary Search tree.

Below is C++ implementation of find minimum operation.

// A C++ program to demonstrate find minimum on KD tree
#include <bits/stdc++.h>
using namespace std;
const int k = 2;
// A structure to represent node of kd tree
struct Node {
    int point[k]; // To store k dimensional point
    Node *left, *right;
// A method to create a node of K D tree
struct Node* newNode(int arr[])
    struct Node* temp = new Node;
    for (int i = 0; i < k; i++)
        temp->point[i] = arr[i];
    temp->left = temp->right = NULL;
    return temp;
// Inserts a new node and returns root of modified tree
// The parameter depth is used to decide axis of comparison
Node* insertRec(Node* root, int point[], unsigned depth)
    // Tree is empty?
    if (root == NULL)
        return newNode(point);
    // Calculate current dimension (cd) of comparison
    unsigned cd = depth % k;
    // Compare the new point with root on current dimension 'cd'
    // and decide the left or right subtree
    if (point[cd] < (root->point[cd]))
        root->left = insertRec(root->left, point, depth + 1);
        root->right = insertRec(root->right, point, depth + 1);
    return root;
// Function to insert a new point with given point in
// KD Tree and return new root. It mainly uses above recursive
// function "insertRec()"
Node* insert(Node* root, int point[])
    return insertRec(root, point, 0);
// A utility function to find minimum of three integers
int min(int x, int y, int z)
    return min(x, min(y, z));
// Recursively finds minimum of d'th dimension in KD tree
// The parameter depth is used to determine current axis.
int findMinRec(Node* root, int d, unsigned depth)
    // Base cases
    if (root == NULL)
        return INT_MAX;
    // Current dimension is computed using current depth and total
    // dimensions (k)
    unsigned cd = depth % k;
    // Compare point with root with respect to cd (Current dimension)
    if (cd == d) {
        if (root->left == NULL)
            return root->point[d];
        return min(root->point[d], findMinRec(root->left, d, depth + 1));
    // If current dimension is different then minimum can be anywhere
    // in this subtree
    return min(root->point[d],
               findMinRec(root->left, d, depth + 1),
               findMinRec(root->right, d, depth + 1));
// A wrapper over findMinRec(). Returns minimum of d'th dimension
int findMin(Node* root, int d)
    // Pass current level or depth as 0
    return findMinRec(root, d, 0);
// Driver program to test above functions
int main()
    struct Node* root = NULL;
    int points[][k] = { { 30, 40 }, { 5, 25 },
       { 70, 70 }, { 10, 12 }, { 50, 30 }, { 35, 45 } };
    int n = sizeof(points) / sizeof(points[0]);
    for (int i = 0; i < n; i++)
        root = insert(root, points[i]);
    cout << "Minimum of 0'th dimension is " << findMin(root, 0) << endl;
    cout << "Minimum of 1'th dimension is " << findMin(root, 1) << endl;
    return 0;


Minimum of 0'th dimension is 5
Minimum of 1'th dimension is 12


This article is contributed by Ashish Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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