二叉树问题中的求解二叉树所有路径的C语言代码

以下是求解二叉树所有路径的C语言代码：

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

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

void printPaths(struct TreeNode* root, int path[], int pathLen) {
    if (!root) return;
    path[pathLen++] = root->val;
    if (!root->left && !root->right) {
        for (int i = 0; i < pathLen; i++) {
            printf("%d ", path[i]);
        }
        printf("\n");
    } else {
        printPaths(root->left, path, pathLen);
        printPaths(root->right, path, pathLen);
    }
}

int main() {
    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 = NULL;
    root->left->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 = 4;
    root->right->left->left = NULL;
    root->right->left->right = NULL;
    root->right->right = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root->right->right->val = 5;
    root->right->right->left = NULL;
    root->right->right->right = NULL;
    int path[100];
    printPaths(root, path, 0);
    return 0;
}

这段代码实现了二叉树所有路径的打印，其中 `printPaths` 函数采用递归实现，每次递归时将当前节点的值加入到 `path` 数组中，并判断当前节点是否为叶子节点，若为叶子节点，则打印 `path` 数组中的所有值，否则继续递归遍历其左右子节点。 `main` 函数中构建了一个二叉树并调用 `printPaths` 函数进行遍历打印。