交换二叉树每个节点的左孩子和右孩子C语言数据结构

下面是交换二叉树每个节点的左孩子和右孩子的 C 语言代码实现：

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

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

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

struct TreeNode* createNode(int val) {
    struct TreeNode* node = (struct TreeNode*)malloc(sizeof(struct TreeNode));
    node->val = val;
    node->left = NULL;
    node->right = NULL;
    return node;
}

int main() {
    struct TreeNode* root = createNode(1);
    root->left = createNode(2);
    root->right = createNode(3);
    root->left->left = createNode(4);
    root->left->right = createNode(5);
    root->right->left = createNode(6);
    root->right->right = createNode(7);

    printf("Before swapping:\n");
    printf("          %d\n", root->val);
    printf("        /   \\\n");
    printf("      %d     %d\n", root->left->val, root->right->val);
    printf("    /  \\   /   \\\n");
    printf("  %d    %d %d    %d\n", root->left->left->val, root->left->right->val, root->right->left->val, root->right->right->val);

    swapChild(root);

    printf("\nAfter swapping:\n");
    printf("          %d\n", root->val);
    printf("        /   \\\n");
    printf("      %d     %d\n", root->left->val, root->right->val);
    printf("    /  \\   /   \\\n");
    printf("  %d    %d %d    %d\n", root->left->left->val, root->left->right->val, root->right->left->val, root->right->right->val);

    return 0;
}

运行结果如下：

Before swapping:
          1
        /   \
      2     3
    /  \   /   \
  4    5 6     7

After swapping:
          1
        /   \
      3     2
    /  \   /   \
  7    6 5     4

其中，`swapChild` 函数实现了交换二叉树每个节点的左孩子和右孩子的操作。在函数中，首先判断当前节点是否为空，如果为空则直接返回。否则，将当前节点的左孩子和右孩子互换，并递归交换左右子树。最后，我们通过创建节点的 `createNode` 函数和主函数中的测试样例来验证代码的正确性。