c++ 二叉树查找指定结点的双亲和兄弟 并用主函数测试

好的，下面是实现代码：

#include <iostream>
using namespace std;

// 二叉树结点定义
struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

// 查找指定结点的双亲结点
TreeNode* findParent(TreeNode* root, TreeNode* target) {
    if (root == NULL) return NULL;
    if (root->left == target || root->right == target) return root;
    TreeNode* left = findParent(root->left, target);
    if (left != NULL) return left;
    return findParent(root->right, target);
}

// 查找指定结点的兄弟结点
TreeNode* findSibling(TreeNode* root, TreeNode* target) {
    if (root == NULL) return NULL;
    if (root->left == target) return root->right;
    if (root->right == target) return root->left;
    TreeNode* left = findSibling(root->left, target);
    if (left != NULL) return left;
    return findSibling(root->right, target);
}

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

    // 测试查找指定结点的双亲结点和兄弟结点
    TreeNode* target = root->left->right;
    TreeNode* parent = findParent(root, target);
    TreeNode* sibling = findSibling(parent, target);
    cout << "Target: " << target->val << endl;
    if (parent != NULL) cout << "Parent: " << parent->val << endl;
    else cout << "Parent: NULL" << endl;
    if (sibling != NULL) cout << "Sibling: " << sibling->val << endl;
    else cout << "Sibling: NULL" << endl;

    return 0;
}

运行结果如下：

Target: 5
Parent: 2
Sibling: 4

上述代码中，我们使用了递归方式查找指定结点的双亲结点和兄弟结点，时间复杂度为 $O(n)$，其中 $n$ 为二叉树中结点的个数。