题解 | #翻转牛群结构#

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param root TreeNode类 
 * @return TreeNode类
 */
struct TreeNode* invertTree(struct TreeNode* root ) {
    // write code here
    if(root == NULL)
    return root;
    struct TreeNode* left = invertTree(root->left);
   //即left = root->le
    struct TreeNode* right = invertTree(root->right);
  //即right = root->right;
  
    root->left = right;
    root->right = left;
    return root;
  //返回的是根节点
}