采用二叉链表存储结构，交换二叉树中每个结点的左孩子与右孩子，用c++

好的，下面是交换二叉树中每个结点的左孩子与右孩子的C++代码实现：

#include<iostream>
using namespace std;

//定义二叉树结点
struct TreeNode{
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x):val(x),left(NULL),right(NULL){}
};

//交换二叉树中每个结点的左孩子与右孩子
TreeNode* invertTree(TreeNode* root) {
    if(root == NULL) return NULL;
    TreeNode* temp = root->left;
    root->left = invertTree(root->right);
    root->right = invertTree(temp);
    return root;
}

//中序遍历二叉树
void inorderTraversal(TreeNode* root){
    if(root == NULL) return;
    inorderTraversal(root->left);
    cout<<root->val<<" ";
    inorderTraversal(root->right);
}

int main(){
    //构建二叉树
    TreeNode* root = new TreeNode(4);
    root->left = new TreeNode(2);
    root->right = new TreeNode(7);
    root->left->left = new TreeNode(1);
    root->left->right = new TreeNode(3);
    root->right->left = new TreeNode(6);
    root->right->right = new TreeNode(9);

    //中序遍历原始二叉树
    cout<<"Original tree: ";
    inorderTraversal(root);
    cout<<endl;

    //交换二叉树中每个结点的左孩子与右孩子
    root = invertTree(root);

    //中序遍历交换后的二叉树
    cout<<"Inverted tree: ";
    inorderTraversal(root);
    cout<<endl;

    return 0;
}

运行结果：

Original tree: 1 2 3 4 6 7 9 
Inverted tree: 9 7 6 4 3 2 1

希望能够帮助到你！