我的目标是生成600,000个数字并将其插入AVL树中。然后，在从AVL树中删除一个号码之后，我应该重新洗牌。
现在，我面临两个问题。
1。我想在我的主要方法中使用for循环来插入数字。但是，它始终生成并打印一个数字。2。如何重新排列这些数字？我认为打印所有数字并将它们放入列表中是可行的，然后重新洗牌，这有点复杂。
这是我的代码（基于GeeksforGeeks）
方法部分：]
static class Node {
int key, height;
Node left, right;
Node(int d) {
key = d;
height = 1;
}
}
static class AVLTree {
Node root;
// A utility function to get the height of the tree
int height(Node N) {
if (N == null)
return 0;
return N.height;
}
// A utility function to get maximum of two integers
int max(int a, int b) {
return (a > b) ? a : b;
}
// A utility function to right rotate subtree rooted with y
// See the diagram given above.
Node rightRotate(Node y) {
Node x = y.left;
Node T2 = x.right;
// Perform rotation
x.right = y;
y.left = T2;
// Update heights
y.height = max(height(y.left), height(y.right)) + 1;
x.height = max(height(x.left), height(x.right)) + 1;
// Return new root
return x;
}
// A utility function to left rotate subtree rooted with x
// See the diagram given above.
Node leftRotate(Node x) {
Node y = x.right;
Node T2 = y.left;
// Perform rotation
y.left = x;
x.right = T2;
// Update heights
x.height = max(height(x.left), height(x.right)) + 1;
y.height = max(height(y.left), height(y.right)) + 1;
// Return new root
return y;
}
// Get Balance factor of node N
int getBalance(Node N) {
if (N == null)
return 0;
return height(N.left) - height(N.right);
}
Node insert(Node node, int key) {
/* 1. Perform the normal BST insertion */
if (node == null)
return (new Node(key));
if (key < node.key)
node.left = insert(node.left, key);
else if (key > node.key)
node.right = insert(node.right, key);
else // Duplicate keys not allowed
return node;
/* 2. Update height of this ancestor node */
node.height = 1 + max(height(node.left),
height(node.right));
/* 3. Get the balance factor of this ancestor
node to check whether this node became
unbalanced */
int balance = getBalance(node);
// If this node becomes unbalanced, then there
// are 4 cases Left Left Case
if (balance > 1 && key < node.left.key)
return rightRotate(node);
// Right Right Case
if (balance < -1 && key > node.right.key)
return leftRotate(node);
// Left Right Case
if (balance > 1 && key > node.left.key) {
node.left = leftRotate(node.left);
return rightRotate(node);
}
// Right Left Case
if (balance < -1 && key < node.right.key) {
node.right = rightRotate(node.right);
return leftRotate(node);
}
/* return the (unchanged) node pointer */
return node;
}
Node minValueNode(Node node)
{
Node current = node;
/* loop down to find the leftmost leaf */
while (current.left != null)
current = current.left;
return current;
}
Node deleteNode(Node root, int key)
{
// STEP 1: PERFORM STANDARD BST DELETE
if (root == null)
return root;
// If the key to be deleted is smaller than
// the root's key, then it lies in left subtree
if (key < root.key)
root.left = deleteNode(root.left, key);
// If the key to be deleted is greater than the
// root's key, then it lies in right subtree
else if (key > root.key)
root.right = deleteNode(root.right, key);
// if key is same as root's key, then this is the node
// to be deleted
else
{
// node with only one child or no child
if ((root.left == null) || (root.right == null))
{
Node temp = null;
if (temp == root.left)
temp = root.right;
else
temp = root.left;
// No child case
if (temp == null)
{
temp = root;
root = null;
}
else // One child case
root = temp; // Copy the contents of
// the non-empty child
}
else
{
// node with two children: Get the inorder
// successor (smallest in the right subtree)
Node temp = minValueNode(root.right);
// Copy the inorder successor's data to this node
root.key = temp.key;
// Delete the inorder successor
root.right = deleteNode(root.right, temp.key);
}
}
// If the tree had only one node then return
if (root == null)
return root;
// STEP 2: UPDATE HEIGHT OF THE CURRENT NODE
root.height = max(height(root.left), height(root.right)) + 1;
// STEP 3: GET THE BALANCE FACTOR OF THIS NODE (to check whether
// this node became unbalanced)
int balance = getBalance(root);
// If this node becomes unbalanced, then there are 4 cases
// Left Left Case
if (balance > 1 && getBalance(root.left) >= 0)
return rightRotate(root);
// Left Right Case
if (balance > 1 && getBalance(root.left) < 0)
{
root.left = leftRotate(root.left);
return rightRotate(root);
}
// Right Right Case
if (balance < -1 && getBalance(root.right) <= 0)
return leftRotate(root);
// Right Left Case
if (balance < -1 && getBalance(root.right) > 0)
{
root.right = rightRotate(root.right);
return leftRotate(root);
}
return root;
}
// A utility function to print preorder traversal
// of the tree.
// The function also prints height of every node
void preOrder(Node node) {
if (node != null) {
System.out.print(node.key + " ");
preOrder(node.left);
preOrder(node.right);
}
}
}
主要方法] >>
public static void main(String[] args) { // TODO Auto-generated method stub AVLTree tree = new AVLTree(); int rand = (int) (Math.random()*600000)+1; /* Constructing tree given in the above figure */ for (int i=0;i<600000;i++) { tree.root=tree.insert(tree.root,rand); } System.out.println("Preorder traversal" + " of constructed tree is : "); tree.preOrder(tree.root); }非常感谢！
我的目标是生成600,000个数字并将其插入AVL树中。然后，在从AVL树中删除一个号码之后，我应该重新洗牌。现在我面临两个问题。 1.我想使用...
Nomadmaker是正确的，您正在声明时生成随机数，并且每次都在for循环中使用该随机数，您应该尝试此操作