226. Invert Binary Tree

Invert a binary tree.

     4
   /   \   2     7
 / \   / \ 1   3 6   9

to

     4
   /   \   7     2
 / \   / \ 9   6 3   1

Trivia:
This problem was inspired by this original tweet by Max Howell:

Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck off.

算是一件趣事吧
Howell can’t invert a binary tree on a whiteboard so fuck off.
我想说，我也不会写，哈哈

//这是递归的写法，通俗易懂
class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {

    if(root){

      invertTree(root->left);
      invertTree(root->right);

      swap(root->right, root->left);
    }

         return root;
    }
};

//这是非递归的写法，其实也是用递归的机理，利用栈的先进后出的特性，一次一次弹出来

TreeNode* invertTree(TreeNode* root) {
    std::stack<TreeNode*> stk;
    stk.push(root);

    while (!stk.empty()) {
        TreeNode* p = stk.top();
        stk.pop();
        if (p) {
            stk.push(p->left);
            stk.push(p->right);
            std::swap(p->left, p->right);
        }
    }
    return root;
}

在提交的时候发现，不能用root.left，要用root->left，才能通过
是因参数是 *root，如果参数是root就可以用 “.”了吗？改了半天

class Solution { public: TreeNode invertTree(TreeNode root) { if(root == NULL) return 0; if(root) invertTree(root.left); invertTree(root.right); swap(root.right, root.left); return root; } }; 

这个函数是无法通过的，

Line 14: no match for ‘operator==’ (operand types are ‘TreeNode’ and ‘long int’)

关于. * -> 树等问题还要好好研究一下 挖个坑先