用c语言写代码输出二叉树中从每个叶子结点到根结点的路径

好的，让我来回答你的问题。以下是用C语言写代码输出二叉树中从每个叶子结点到根结点的路径的步骤：

1. 定义二叉树结构体，包括左右子节点和值三个成员。
2. 定义一个栈结构体，用来存储路径中的节点信息。
3. 定义一个遍历函数，以每个叶子节点为起点进行遍历。
4. 在遍历函数中，使用递归的方式遍历每个节点，当遍历到叶子节点时，将路径上的节点信息压入栈中。
5. 遍历结束后，从栈顶开始输出路径上的节点信息，即可得到从叶子节点到根节点的路径信息。

代码示例：

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

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

struct Stack {
    struct TreeNode *node;
    struct Stack *next;
};

void push(struct Stack **s, struct TreeNode *node) {
    struct Stack *temp = (struct Stack*) malloc(sizeof(struct Stack));
    temp->node = node;
    temp->next = *s;
    *s = temp;
}

struct TreeNode *pop(struct Stack **s) {
    struct Stack *temp = *s;
    struct TreeNode *node = temp->node;
    *s = temp->next;
    free(temp);
    return node;
}

void printPath(struct Stack *s) {
    while (s != NULL) {
        printf("%d ", s->node->val);
        s = s->next;
    }
    printf("\n");
}

void traverse(struct TreeNode *root, struct Stack *s) {
    if (root == NULL) return;
    push(&s, root);
    if (root->left == NULL && root->right == NULL) {
        printPath(s);
    } else {
        traverse(root->left, s);
        traverse(root->right, s);
    }
    pop(&s);
}

int main() {
    // 构建二叉树
    struct TreeNode *root = (struct TreeNode*) malloc(sizeof(struct TreeNode));
    root->val = 1;
    root->left = (struct TreeNode*) malloc(sizeof(struct TreeNode));
    root->left->val = 2;
    root->left->left = (struct TreeNode*) malloc(sizeof(struct TreeNode));
    root->left->left->val = 4;
    root->left->left->left = NULL;
    root->left->left->right = NULL;
    root->left->right = (struct TreeNode*) malloc(sizeof(struct TreeNode));
    root->left->right->val = 5;
    root->left->right->left = NULL;
    root->left->right->right = NULL;
    root->right = (struct TreeNode*) malloc(sizeof(struct TreeNode));
    root->right->val = 3;
    root->right->left = (struct TreeNode*) malloc(sizeof(struct TreeNode));
    root->right->left->val = 6;
    root->right->left->left = NULL;
    root->right->left->right = NULL;
    root->right->right = (struct TreeNode*) malloc(sizeof(struct TreeNode));
    root->right->right->val = 7;
    root->right->right->left = NULL;
    root->right->right->right = NULL;

    // 遍历二叉树
    struct Stack *s = NULL;
    traverse(root, s);

    return 0;
}

输出结果：

4 2 1
5 2 1
6 3 1
7 3 1