※ Leetcode - Tree - 226. Invert Binary Tree（反转二叉树 使用二级指针交换两个指针的地址）

1. Problem Description

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.

 

2. My solution（0ms）

反转二叉树，题目中给出的TreeNode类实现：

/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
思路很简单，递归交换左右子树，关键在于，如何交换两个指针的地址。

如果要交换两个指针的值，需要使用二级指针。

　　void swapp(TreeNode** t1,TreeNode** t2)
{
TreeNode* tmp;
tmp=*t1;
*t1=*t2;
*t2=tmp;
}
My AC code： 

class Solution
{
public:
void swapp(TreeNode** t1,TreeNode** t2)
{
TreeNode* tmp;
tmp=*t1;
*t1=*t2;
*t2=tmp;
}
void DFS(TreeNode* root)
{
if(root!=NULL)
{
DFS(root->left);
DFS(root->right);
swapp(&(root->left),&(root->right));
}
}
TreeNode* invertTree(TreeNode* root)
{
DFS(root);
return root;
}
};

3. Simple solution 

class Solution
{
public:
void DFS(TreeNode* root)
{
if(root!=NULL)
{
DFS(root->left);
DFS(root->right);
swap(root->left,root->right);
}
}
TreeNode* invertTree(TreeNode* root)
{
DFS(root);
return root;
}
};