优化以下代码#include <stdio.h> #include <stdlib.h> typedef struct TreeNode { char data; struct TreeNode* left; struct TreeNode* right; } TreeNode; TreeNode* createTree() { char ch; TreeNode* root; scanf("%c", &ch); if (ch == '#') { return NULL; } root = (TreeNode*)malloc(sizeof(TreeNode)); root->data = ch; root->left = createTree(); root->right = createTree(); return root; } void digui(TreeNode* root) { if (root == NULL) { return; } digui(root->left); printf("%c ", root->data); digui(root->right); } typedef struct StackNode { TreeNode* tree; struct StackNode* next; } StackNode; typedef struct Stack { StackNode* top; int size; } Stack; Stack* createStack() { Stack* stack = (Stack*)malloc(sizeof(Stack)); stack->top = NULL; stack->size = 0; return stack; } void push(Stack* stack, TreeNode* tree) { StackNode* node; node = (StackNode*)malloc(sizeof(StackNode)); node->tree = tree; node->next = stack->top; stack->top = node; stack->size++; } TreeNode* pop(Stack* stack) { TreeNode* tree; StackNode* temp; if (stack->size == 0) { return NULL; } tree = stack->top->tree; temp = stack->top; stack->top = stack->top->next; stack->size--; free(temp); return tree; } void feidigui(TreeNode* root) { Stack* stack; TreeNode* p; stack = createStack(); p = root; while (p != NULL || stack->size != 0) { while (p != NULL) { push(stack, p); p = p->left; } if (stack->size != 0) { p = pop(stack); printf("%c ", p->data); p = p->right; } } } int getHeight(TreeNode* root) { int leftHeight,rightHeight,max; if (root == NULL) { return 0; } leftHeight = getHeight(root->left); rightHeight = getHeight(root->right); max=leftHeight>rightHeight?leftHeight:rightHeight; return max+1; }

首先，这段代码的缩进不太规范，可以调整一下，使得代码更加清晰易读。

其次，在函数中出现了多次重复的代码，如创建栈、入栈、出栈等，可以将这些代码封装成一个通用的函数，以便在程序中反复使用。

最后，可以考虑使用更加简洁的写法，如使用三目运算符代替if-else语句，使用递归函数代替while循环等，以提高代码的可读性和可维护性。同时，可以加上注释，方便他人理解代码逻辑。

下面是一个优化后的代码示例：

#include <stdio.h> 
#include <stdlib.h> 

typedef struct TreeNode { 
    char data; 
    struct TreeNode *left; 
    struct TreeNode *right; 
} TreeNode; 

// 创建二叉树 
TreeNode* createTree() { 
    char ch; 
    scanf("%c", &ch); 
    if (ch == '#') { 
        return NULL; 
    } 
    TreeNode *root = (TreeNode*)malloc(sizeof(TreeNode)); 
    root->data = ch; 
    root->left = createTree(); 
    root->right = createTree(); 
    return root; 
} 

// 递归遍历二叉树 
void digui(TreeNode *root) { 
    if (root == NULL) { 
        return; 
    } 
    digui(root->left); 
    printf("%c ", root->data); 
    digui(root->right); 
} 

typedef struct StackNode { 
    TreeNode *tree; 
    struct StackNode *next; 
} StackNode; 

typedef struct Stack { 
    StackNode *top; 
    int size; 
} Stack; 

// 创建栈 
Stack* createStack() { 
    Stack *stack = (Stack*)malloc(sizeof(Stack)); 
    stack->top = NULL; 
    stack->size = 0; 
    return stack; 
} 

// 入栈 
void push(Stack *stack, TreeNode *tree) { 
    StackNode *node = (StackNode*)malloc(sizeof(StackNode)); 
    node->tree = tree; 
    node->next = stack->top; 
    stack->top = node; 
    stack->size++; 
} 

// 出栈 
TreeNode* pop(Stack *stack) { 
    if (stack->size == 0) { 
        return NULL; 
    } 
    TreeNode *tree = stack->top->tree; 
    StackNode *temp = stack->top; 
    stack->top = stack->top->next; 
    stack->size--; 
    free(temp); 
    return tree; 
} 

// 非递归遍历二叉树 
void feidigui(TreeNode *root) { 
    Stack *stack = createStack(); 
    TreeNode *p = root; 
    while (p != NULL || stack->size != 0) { 
        if (p != NULL) { 
            push(stack, p); 
            p = p->left; 
        } else { 
            p = pop(stack); 
            printf("%c ", p->data); 
            p = p->right; 
        } 
    } 
} 

// 获取二叉树的高度 
int getHeight(TreeNode *root) { 
    if (root == NULL) { 
        return 0; 
    } 
    int leftHeight = getHeight(root->left); 
    int rightHeight = getHeight(root->right); 
    return leftHeight > rightHeight ? leftHeight + 1 : rightHeight + 1; 
} 

int main() { 
    TreeNode *root = createTree(); 
    digui(root); 
    printf("\n"); 
    feidigui(root); 
    printf("\n"); 
    printf("%d\n", getHeight(root)); 
    return 0; 
}