LeetCode 226. Invert Binary Tree--反转二叉树--C++,Python解法--递归，迭代做法

题目地址：Invert Binary Tree - LeetCode

Invert a binary tree.

Example:

Input:

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

Output:

     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 f*** off.

这道题目是经典的反转二叉树，用递归来做非常快。
Python解法如下：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution:def invertTree(self, root: TreeNode) -> TreeNode:if root==None:returnroot.left,root.right=root.right,root.leftself.invertTree(root.left)self.invertTree(root.right)return root

迭代的做法麻烦很多，迭代有BFS和DFS共2种做法，DFS解法如下。
Python解法如下：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution:def invertTree(self, root: TreeNode) -> TreeNode:stack = [root]while stack:node = stack.pop()if node:node.left, node.right = node.right, node.leftstack.extend([node.right, node.left])return root

C++解法如下：

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/
class Solution
{public:TreeNode *invertTree(TreeNode *root){if (root == nullptr){return root;}TreeNode *temp = root->left;root->left = root->right;root->right = temp;invertTree(root->left);invertTree(root->right);return root;}
};

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