调用leftRotate后AVL树打印函数输出异常，求错误定位

我实现了一个打印二叉搜索树（BST）结构的print2D函数，不调用leftRotate时输出正常，但调用该函数后，输出的树结构错误且丢失部分节点。以下是完整的C语言程序：

// C program to insert a node in AVL tree
#include<stdio.h>
#include<stdlib.h>
#define COUNT 10

// 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 the height of the tree
int height(struct Node *N)
{
    if (N == NULL)
        return 0;
    return 1+max(height(N->left), height(N->right));
}

// 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 = 0; // 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 = height(y);
    x->height = height(x);

    // 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 = height(x);
    y->height = height(y);

    // 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);
}

// Recursive function to insert a key in the subtree rooted
// with node and returns the new root of the subtree.
struct Node* insert(struct Node* node, int key)
{
    /* 1. Perform the normal BST insertion */
    if (node == NULL)
        return(newNode(key));

    if (key < node->key)
        node->left = insert(node->left, key);
    else if (key > node->key)
        node->right = insert(node->right, key);
    else // Equal keys are not allowed in BST
        return node;

    /* 2. Update height of this ancestor node */
    node->height = height(node);

    /* 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;
}

// 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)
    {
        printf("%d ", root->key);
        preOrder(root->left);
        preOrder(root->right);
    }
}


void print2DUtil(struct Node*root, int space)
{
    // Base case
    if (root == NULL)
        return;
 
    // Increase distance between levels
    space += COUNT;
 
    // Process right child first
    print2DUtil(root->right, space);
 
    // Print current node after space
    // count
    printf("\n");
    for (int i = COUNT; i < space; i++)
        printf(" ");
    printf("%d\n", root->key);
 
    // Process left child
    print2DUtil(root->left, space);
}

// Wrapper over print2DUtil()
void print2D(struct Node*root)
{
   // Pass initial space count as 0
   print2DUtil(root, 0);
}
 

/* Driver program to test above function*/
int main()
{
struct Node *root = NULL;
    root = insert(root, 20);
    root = insert(root, 11);
    root = insert(root, 32);
    root = insert(root, 4);
    root = insert(root, 16);
    root = insert(root, 25);
    root = insert(root, 36);
    root = insert(root, 3);
    root = insert(root, 7);
    root = insert(root, 13);
    root = insert(root, 18);
    root = insert(root, 21);
    root = insert(root, 28);
    root = insert(root, 33);
    root = insert(root, 39);
    

/* The constructed AVL Tree would be
            30
        / \
        20 40
        / \  \
    10 25 50
*/
root = leftRotate(root);
print2D(root);

return 0;
}

问题根源 #

height函数的计算逻辑与newNode中的高度初始化值不匹配，导致旋转后节点高度更新错误，进而破坏了树的结构关系。

• 原height函数中，空节点返回0，叶子节点的计算高度为1 + max(0,0) = 1
• 但newNode中初始化node->height = 0，与计算逻辑矛盾

当执行leftRotate时，调用height()更新节点高度，递归计算过程中会出现高度值混乱，最终导致打印函数遍历树时出现异常，表现为结构错误、节点丢失。

修复方案 #

任选以下一种方式即可解决：

方式一：修正newNode的高度初始化值 #

将叶子节点的初始高度设为1，匹配height函数的计算逻辑：

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; // 修改为1
    return(node);
}

方式二：修正height函数的逻辑 #

让空节点返回-1，这样叶子节点的计算高度为1 + max(-1,-1) = 0，匹配初始值：

int height(struct Node *N)
{
    if (N == NULL)
        return -1; // 修改为-1
    return 1+max(height(N->left), height(N->right));
}

验证 #

修复后执行leftRotate再调用print2D，即可正确输出旋转后的树结构，节点不会丢失，树的层级关系也能准确展示。

内容的提问来源于stack exchange，提问作者ferocioussprouts122