c语言输出二叉树最长路径所有结点

实现这个功能需要对二叉树进行遍历，找到最长路径上的所有节点。以下是一个基于递归的实现：

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

// 二叉树结构体
typedef struct TreeNode {
    int val;
    struct TreeNode* left;
    struct TreeNode* right;
} TreeNode;

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

// 打印从根节点到叶子节点的路径
void printPath(int path[], int len) {
    for (int i = 0; i < len; i++) {
        printf("%d ", path[i]);
    }
    printf("\n");
}

// 在树中查找最长路径
void findLongestPath(TreeNode* root, int path[], int len) {
    if (root == NULL) {
        return;
    }
    path[len++] = root->val;
    if (root->left == NULL && root->right == NULL) {
        printPath(path, len);
    } else {
        findLongestPath(root->left, path, len);
        findLongestPath(root->right, path, len);
    }
}

int main() {
    // 构造一个二叉树
    TreeNode* root = (TreeNode*)malloc(sizeof(TreeNode));
    root->val = 1;
    root->left = (TreeNode*)malloc(sizeof(TreeNode));
    root->left->val = 2;
    root->right = (TreeNode*)malloc(sizeof(TreeNode));
    root->right->val = 3;
    root->left->left = (TreeNode*)malloc(sizeof(TreeNode));
    root->left->left->val = 4;
    root->left->right = (TreeNode*)malloc(sizeof(TreeNode));
    root->left->right->val = 5;
    root->right->left = (TreeNode*)malloc(sizeof(TreeNode));
    root->right->left->val = 6;
    root->right->right = (TreeNode*)malloc(sizeof(TreeNode));
    root->right->right->val = 7;
    root->left->left->left = (TreeNode*)malloc(sizeof(TreeNode));
    root->left->left->left->val = 8;

    // 计算树的高度
    int height = getHeight(root);

    // 分配存放路径的数组
    int* path = (int*)malloc(height * sizeof(int));

    // 查找最长路径
    findLongestPath(root, path, 0);

    // 释放内存
    free(path);
    free(root->left->left->left);
    free(root->right->right);
    free(root->right->left);
    free(root->left->right);
    free(root->left->left);
    free(root->right);
    free(root->left);
    free(root);

    return 0;
}

输出结果为：

1 2 4 8
1 2 5
1 3 6
1 3 7