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K Dimensional Tree | Set 3 (Delete)

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  • Difficulty Level : Hard
  • Last Updated : 18 Jul, 2022
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We strongly recommend to refer below posts as a prerequisite of this.
K Dimensional Tree | Set 1 (Search and Insert) 
K Dimensional Tree | Set 2 (Find Minimum)
In this post delete is discussed. The operation is to delete a given point from K D Tree.

Like Binary Search Tree Delete, we recursively traverse down and search for the point to be deleted. Below are steps are followed for every node visited.

1) If current node contains the point to be deleted 

  1. If node to be deleted is a leaf node, simply delete it (Same as BST Delete)
  2. If node to be deleted has right child as not NULL (Different from BST) 
    1. Find minimum of current node’s dimension in right subtree.
    2. Replace the node with above found minimum and recursively delete minimum in right subtree.
  3. Else If node to be deleted has left child as not NULL (Different from BST) 
    1. Find minimum of current node’s dimension in left subtree.
    2. Replace the node with above found minimum and recursively delete minimum in left subtree.
    3. Make new left subtree as right child of current node.

2) If current doesn’t contain the point to be deleted  

  1. If node to be deleted is smaller than current node on current dimension, recur for left subtree.
  2. Else recur for right subtree.

Why 1.b and 1.c are different from BST? 
In BST delete, if a node’s left child is empty and right is not empty, we replace the node with right child. In K D Tree, doing this would violate the KD tree property as dimension of right child of node is different from node’s dimension. For example, if node divides point by x axis values. then its children divide by y axis, so we can’t simply replace node with right child. Same is true for the case when right child is not empty and left child is empty.

Why 1.c doesn’t find max in left subtree and recur for max like 1.b? 
Doing this violates the property that all equal values are in right subtree. For example, if we delete (!0, 10) in below subtree and replace if with 

Wrong Way (Equal key in left subtree after deletion)
            (5, 6)                             (4, 10)
             /              Delete(5, 6)         /  
        (4, 10)            ------------>     (4, 20)
           (4, 20) 

Right way (Equal key in right subtree after deletion)
             (5, 6)                          (4, 10)
             /              Delete(5, 6)           \
         (4, 10)            ------------>         (4, 20)
             (4, 20) 

Example of Delete: 
Delete (30, 40): Since right child is not NULL and dimension of node is x, we find the node with minimum x value in right child. The node is (35, 45), we replace (30, 40) with (35, 45) and delete (30, 40).


Delete (70, 70): Dimension of node is y. Since right child is NULL, we find the node with minimum y value in left child. The node is (50, 30), we replace (70, 70) with (50, 30) and recursively delete (50, 30) in left subtree. Finally we make the modified left subtree as right subtree of (50, 30).  


Below is C++ implementation of K D Tree delete.


// A C++ program to demonstrate delete in K D tree
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
Node *minNode(Node *x, Node *y, Node *z, int d)
    Node *res = x;
    if (y != NULL && y->point[d] < res->point[d])
       res = y;
    if (z != NULL && z->point[d] < res->point[d])
       res = z;
    return res;
// Recursively finds minimum of d'th dimension in KD tree
// The parameter depth is used to determine current axis.
Node *findMinRec(Node* root, int d, unsigned depth)
    // Base cases
    if (root == NULL)
        return NULL;
    // 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;
        return findMinRec(root->left, d, depth+1);
    // If current dimension is different then minimum can be anywhere
    // in this subtree
    return minNode(root,
               findMinRec(root->left, d, depth+1),
               findMinRec(root->right, d, depth+1), d);
// A wrapper over findMinRec(). Returns minimum of d'th dimension
Node *findMin(Node* root, int d)
    // Pass current level or depth as 0
    return findMinRec(root, d, 0);
// A utility method to determine if two Points are same
// in K Dimensional space
bool arePointsSame(int point1[], int point2[])
    // Compare individual pointinate values
    for (int i = 0; i < k; ++i)
        if (point1[i] != point2[i])
            return false;
    return true;
// Copies point p2 to p1
void copyPoint(int p1[], int p2[])
   for (int i=0; i<k; i++)
       p1[i] = p2[i];
// Function to delete a given point 'point[]' from tree with root
// as 'root'.  depth is current depth and passed as 0 initially.
// Returns root of the modified tree.
Node *deleteNodeRec(Node *root, int point[], int depth)
    // Given point is not present
    if (root == NULL)
        return NULL;
    // Find dimension of current node
    int cd = depth % k;
    // If the point to be deleted is present at root
    if (arePointsSame(root->point, point))
        // 2.b) If right child is not NULL
        if (root->right != NULL)
            // Find minimum of root's dimension in right subtree
            Node *min = findMin(root->right, cd);
            // Copy the minimum to root
            copyPoint(root->point, min->point);
            // Recursively delete the minimum
            root->right = deleteNodeRec(root->right, min->point, depth+1);
        else if (root->left != NULL) // same as above
            Node *min = findMin(root->left, cd);
            copyPoint(root->point, min->point);
            root->right = deleteNodeRec(root->left, min->point, depth+1);
        else // If node to be deleted is leaf node
            delete root;
            return NULL;
        return root;
    // 2) If current node doesn't contain point, search downward
    if (point[cd] < root->point[cd])
        root->left = deleteNodeRec(root->left, point, depth+1);
        root->right = deleteNodeRec(root->right, point, depth+1);
    return root;
// Function to delete a given point from K D Tree with 'root'
 Node* deleteNode(Node *root, int point[])
   // Pass depth as 0
   return deleteNodeRec(root, point, 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]);
    // Delete (30, 40);
    root = deleteNode(root, points[0]);
    cout << "Root after deletion of (30, 40)\n";
    cout << root->point[0] << ", " << root->point[1] << endl;
    return 0;


Root after deletion of (30, 40)
35, 45

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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