226. Invert Binary Tree

细节处理题，递归对左右树进行翻转，
1，注意穿参，因为需要修改左右指针，需要传递指针的地址
em(&root->left,&root->right);

2，注意对单子树翻转的shh

void em(struct TreeNode **left, struct TreeNode **right)
{
    //swap(left->left,right->right) swap(left->right, right->left)
    if(*left == NULL && *right == NULL){
        
    }else if(*left != NULL && *right == NULL){
        *right = *left;
        *left = NULL;
        em(&(*right)->left,&(*right)->right);
    }
    else if(*left == NULL && *right != NULL){
        *left = *right;
        *right = NULL;
        em(&(*left)->left, &(*left)->right);
    }
    else{
        int tmp = (*left)->val;
        (*left)->val = (*right)->val;
        (*right)->val = tmp;
        em(&(*left)->left,&(*right)->right); 
        em(&(*left)->right, &(*right)->left);
    }


}
struct TreeNode* invertTree(struct TreeNode* root) {
    if(root == NULL)
        return NULL;

    em(&root->left,&root->right);

    return root;
}