LeetCode 226. Invert Binary Tree--反轉二叉樹--C++,Python解法

LeetCode 226. Invert Binary Tree–反轉二叉樹–C++,Python解法

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LeetCode題解專欄：LeetCode題解
我做的所有的LeetCode的題目都放在這個專欄裏，大部分題目Java和Python的解法都有。

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題目地址：Invert Binary Tree - LeetCode

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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.

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這道題目是經典的反轉二叉樹，用遞歸來做非常快。
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 = None

class Solution:
    def invertTree(self, root: TreeNode) -> TreeNode:
        if root==None:
            return
        root.left,root.right=root.right,root.left
        self.invertTree(root.left)
        self.invertTree(root.right)
        return root

迭代的做法麻煩很多，迭代有BFS和DFS共2種做法，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 = None

class Solution:
    def invertTree(self, root: TreeNode) -> TreeNode:
            stack = [root]
            while stack:
                node = stack.pop()
                if node:
                    node.left, node.right = node.right, node.left
                    stack.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;
    }
};