226 Invert Binary Tree 翻转二叉树

Description:
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.

题目描述:
翻转一棵二叉树。

示例：

输入：

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

输出：

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

备注:
这个问题是受到 Max Howell的 原问题 启发的 ：

谷歌：我们90％的工程师使用您编写的软件(Homebrew)，但是您却无法在面试时在白板上写出翻转二叉树这道题，这太糟糕了。

思路:

采用LeetCode #107 Binary Tree Level Order Traversal II 二叉树的层次遍历 II类似的思路, 迭代或者递归解决
时间复杂度O(n), 空间复杂度O(n)

代码:
C++:

/**
 * 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) return root;
        queue<TreeNode*> q;
        q.push(root);
        while (!q.empty()) 
        {
            TreeNode* cur = q.front();
            q.pop();
            TreeNode* temp = cur -> right;
            cur -> right = cur -> left;
            cur -> left = temp;
            if (cur -> left) q.push(cur -> left);
            if (cur -> right) q.push(cur -> right);
        }
        return root;
    }
};

Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) return null;
        TreeNode temp = root.left;
        root.left = invertTree(root.right);
        root.right = invertTree(temp);
        return root;
    }
}

Python:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def invertTree(self, root: TreeNode) -> TreeNode:
        if not root:
            return root
        root.left, root.right = self.invertTree(root.right), self.invertTree(root.left)
        return root