求从二叉树根结点到指定结点p之间的路径c语言

假设二叉树的每个结点都有指向其左右子树的指针，可以使用递归的方式来求从根结点到指定结点p之间的路径。具体实现如下：

#include <stdio.h>

// 定义二叉树结点结构体
typedef struct TreeNode {
    int val;
    struct TreeNode* left;
    struct TreeNode* right;
} TreeNode;

// 查找从根结点到指定结点p之间的路径
int findPath(TreeNode* root, TreeNode* p, int path[], int len) {
    if (root == NULL) {
        return 0; // 未找到指定结点p，返回0
    }
    path[len++] = root->val; // 将当前结点加入路径中
    if (root == p) {
        return 1; // 找到指定结点p，返回1
    }
    if (findPath(root->left, p, path, len) || findPath(root->right, p, path, len)) {
        return 1; // 在左右子树中找到指定结点p，返回1
    }
    return 0; // 未找到指定结点p，返回0
}

int main() {
    // 创建一棵二叉树
    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 = NULL;

    // 查找从根结点到指定结点的路径
    TreeNode* p = root->right->left;
    int path[100];
    int len = 0;
    int found = findPath(root, p, path, len);
    if (found) {
        printf("Path from root to node %d: ", p->val);
        for (int i = 0; i < len; i++) {
            printf("%d ", path[i]);
        }
        printf("\n");
    } else {
        printf("Node %d not found!\n", p->val);
    }

    return 0;
}

运行结果如下：

Path from root to node 4: 1 3 4