用c++编写算法将二叉树中所有结点的左右子树相互交换

很高兴回答你的问题！这是一道很经典的二叉树问题。以下是 C 语言代码实现：

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

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

void swapChildren(struct TreeNode* node){
    if(node == NULL) return;
    
    struct TreeNode* temp = node->left;
    node->left = node->right;
    node->right = temp;
    
    swapChildren(node->left);
    swapChildren(node->right);
}

int main() {
    //构造一颗二叉树
    struct TreeNode* root = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root -> val = 1;
    root -> left = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root -> right = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root -> left -> val = 2;
    root -> right -> val = 3;
    root -> left -> left = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root -> left -> right = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root -> left -> left -> val = 4;
    root -> left -> right -> val = 5;
    root -> right -> left = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root -> right -> right = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    root -> right -> left -> val = 6;
    root -> right -> right -> val = 7;

    //交换左右子树
    swapChildren(root);

    //输出交换后的结果
    printf("%d\n", root -> val);
    printf("%d %d\n", root -> left -> val, root -> right -> val);
    printf("%d %d %d %d\n", root -> left -> left -> val, root -> left -> right -> val, root -> right -> left -> val, root -> right -> right -> val);

    return 0;
}

该算法通过递归的方式，遍历二叉树中的每一个节点，并将其左右子树相互交换。时间复杂度为 O(n)，其中 n 为二叉树中节点的个数。