#include<stdio.h>
#include<stdlib.h>
// An AVL tree node
struct node
{
int key;
struct node *left;
struct node *right;
int height;
};
// A utility function to get maximum of two integers
int max(int a, int b);
// A utility function to get height of the tree
int height(struct 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;
}
/* Helper function that allocates a new node with the given key and
NULL left and right pointers. */
struct node* newNode(int key)
{
struct node* node = (struct node*)
malloc(sizeof(struct node));
node->key = key;
node->left = NULL;
node->right = NULL;
node->height = 1; // new node is initially added at leaf
return(node);
}
// A utility function to right rotate subtree rooted with y
// See the diagram given above.
struct node *rightRotate(struct node *y)
{
struct node *x = y->left;
struct 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.
struct node *leftRotate(struct node *x)
{
struct node *y = x->right;
struct 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(struct node *N)
{
if (N == NULL)
return 0;
return height(N->left) - height(N->right);
}
struct node* insert(struct node* node, int key)
{
/* 1. Perform the normal BST rotation */
if (node == NULL)
return(newNode(key));
if (key < node->key)
node->left = insert(node->left, key);
else
node->right = insert(node->right, key);
/* 2. Update height of this ancestor node */
node->height = max(height(node->left), height(node->right)) + 1;
/* 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;
}
/* Given a non-empty binary search tree, return the node with minimum
key value found in that tree. Note that the entire tree does not
need to be searched. */
struct node * minValueNode(struct node* node)
{
struct node* current = node;
/* loop down to find the leftmost leaf */
while (current->left != NULL)
current = current->left;
return current;
}
struct node* deleteNode(struct 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) )
{
struct node *temp = root->left ? root->left : root->right;
// No child case
if(temp == NULL)
{
temp = root;
root = NULL;
}
else // One child case
*root = *temp; // Copy the contents of the non-empty child
free(temp);
}
else
{
// node with two children: Get the inorder successor (smallest
// in the right subtree)
struct 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(struct node *root)
{
if(root != NULL)
{
preOrder(root->left);
printf("%d ", root->key);
preOrder(root->right);
}
}
/* Drier program to test above function*/
int main()
{
struct node *root = NULL;
/* Constructing tree given in the above figure */
root = insert(root, 9);
root = insert(root, 5);
root = insert(root, 10);
root = insert(root, 0);
root = insert(root, 6);
root = insert(root, 11);
root = insert(root, -1);
root = insert(root, 1);
root = insert(root, 2);
root = insert(root, 10);
printf("Pre order traversal of the constructed AVL tree is \n");
preOrder(root);
root = deleteNode(root, 10);
printf("\nPre order traversal after deletion of 10 \n");
preOrder(root);
return 0;
}
当我给出某个位置时，如何打印元素？ 例如：
root = insert(root, 100); (add 100)
root = insert(root, 300); (add 300)
root = insert(root, 200); (add 200)
root = insert(root, 200); (add 400)
PRINT 1 -> Should print the first element (sorted). Which is 100
PRINT 3 -> Should print the third element (sorted). Which is 300
如何在AVL树上实现此PRINT？
我试图修改此功能，但我没有成功
void preOrder(struct node *root)
{
if(root != NULL)
{
preOrder(root->left);
printf("%d ", root->key);
preOrder(root->right);
}
}
我无法在此处使用if语句来检查传递的值，因为它们都是同时传递的。
preOrder函数将对数字进行排序。从低到大。
答案 0 :(得分：0)
我不明白为什么你不能修改preOrder来做到这一点 - 你只需要记住＆＃34;当前的位置在此过程中更新它：
int findPositionPreOrder(
struct node *root,
int targetPos,
int curPos)
{
if(root != NULL)
{
int newPos = findPositionPreOrder(root->left, targetPos, curPos);
newPos++;
if (newPos == targetPos)
{
printf("%d\n", root->key);
}
return findPositionPreOrder(root->right, targetPos, newPos);
}
else
{
return curPos;
}
}
你称之为findPositionPreOrder(root, targetPosition, 0);