用c语言采用非递归方法写一个求二叉树上结点的路径

非常感谢您的问题，这是一道关于二叉树的经典问题。下面是用 C 语言采用非递归方法写一个求二叉树上结点的路径的示例代码：

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

typedef struct TreeNode {
    int val;
    struct TreeNode* left;
    struct TreeNode* right;
} TreeNode;

// 定义栈数据结构
typedef struct Stack {
    TreeNode** nodeArr;
    int top;
    int size;
} Stack;

// 初始化栈
Stack* initStack(int size) {
    Stack* stack = (Stack*)malloc(sizeof(Stack));
    stack->nodeArr = (TreeNode**)calloc(size, sizeof(TreeNode*));
    stack->top = -1;
    stack->size = size;
    return stack;
}

// 判断栈是否为空
bool isStackEmpty(Stack* stack) {
    return stack->top == -1;
}

// 判断栈是否已满
bool isStackFull(Stack* stack) {
    return stack->top == stack->size - 1;
}

// 入栈
bool pushStack(Stack* stack, TreeNode* node) {
    if (isStackFull(stack)) {
        return false;
    }
    stack->nodeArr[++stack->top] = node;
    return true;
}

// 出栈
TreeNode* popStack(Stack* stack) {
    if (isStackEmpty(stack)) {
        return NULL;
    }
    return stack->nodeArr[stack->top--];
}

// 栈顶元素
TreeNode* topStack(Stack* stack) {
    if (isStackEmpty(stack)) {
        return NULL;
    }
    return stack->nodeArr[stack->top];
}

// 二叉树路径遍历
void pathTraversal(TreeNode* root, TreeNode* target) {
    if (root == NULL || target == NULL) {
        return;
    }

    Stack* stack = initStack(100);
    TreeNode* cur = root;
    TreeNode* last = NULL;

    while (cur != NULL || !isStackEmpty(stack)) {
        if (cur) {
            pushStack(stack, cur);
            if (cur == target) {
                printf("找到了目标节点\n");
                printf("路径是：");
                for (int i = 0; i <= stack->top; i++) {
                    printf("%d ", stack->nodeArr[i]->val);
                }
                printf("\n");
                break;
            }
            cur = cur->left;
        } else {
            TreeNode* top = topStack(stack);
            if (top->right && top->right != last) {
                cur = top->right;
            } else {
                last = popStack(stack);
            }
        }
    }
}

int main() {
    TreeNode* root = (TreeNode*)malloc(sizeof(TreeNode));
    root->val = 1;
    root->left = (TreeNode*)malloc(sizeof(TreeNode));
    root->left->val = 2;
    root->left->left = NULL;
    root->left->right = NULL;
    root->right = (TreeNode*)malloc(sizeof(TreeNode));
    root->right->val = 3;
    root->right->left = (TreeNode*)malloc(sizeof(TreeNode));
    root->right->left->val = 4;
    root->right->left->left = NULL;
    root->right->left->right = NULL;
    root->right->right = (TreeNode*)malloc(sizeof(TreeNode));
    root->right->right->val = 5;
    root->right->right->left = NULL;
    root->right->right->right = NULL;

    pathTraversal(root, root->right->left);

    return 0;
}

此代码将从给定的二叉树中查找给定的节点，并输出到根节点的路径。这种方法使用了栈来进行非递归式的遍历，并通过输出栈中所有元素的值来打印路径。

希望这个示例代码可以帮助您解决这个问题，感谢您的提问！