AVL树。按位置打印元素（排序时）

Question

#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;
}

如何在给定某个位置时打印元素？示例：

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

如何在我的 AVL 树上实现这个 PRINT？

我尝试修改这个功能，但没有成功

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 语句来检查正在传递的值，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。

Answer 1

我不明白为什么你不能修改 preOrder 来做到这一点 - 你只需要 "remember" 当前位置并沿途更新它：

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;
    }
}

你可以这样称呼它 findPositionPreOrder(root, targetPosition, 0);

这不是最佳解决方案 - 您可以在找到所需位置后中断递归，而不是遍历树的其余部分。

AVL树。按位置打印元素（排序时）

AVL Tree. Print element by position (when sorted)

algorithm

avl-tree