leetcode 226. Invert Binary Tree 翻转二叉树 python3

时间：2020-6-11

题目地址：https://leetcode-cn.com/problems/invert-binary-tree/

题目难度：Easy

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1：使用递归 - 深度优先遍历（Deep First Search，一杆子到底），先序

递归的两个条件如下：

终止条件：当前节点为null时返回
交换当前节点的左右节点，再递归的交换当前节点的左节点，递归的交换当前节点的右节点

测试case：

• 全二叉树
• 根节点为null
• 左节点为null
• 右节点为null
• 左右节点均为null

代码段1：通过

# 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 None
        if(root.left != None or root.right != None):
            temp = root.left
            root.left = root.right
            root.right = temp
        if(root.left != None):
            self.invertTree(root.left)
        if(root.right != None):
            self.invertTree(root.right)
        return root

总结：

1. 看下优化的代码，递归写的很清爽，下边分别是先序、中序和后序

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

# 先序遍历

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

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

# 中序遍历

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

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

# 后序遍历

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

---

思路2：使用迭代 - 广度优先（Breath First Search，层层遍历）

代码段2：通过

# 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
        queue = [root]
        while(queue):
            temp = queue.pop(0)
            temp.left, temp.right = temp.right, temp.left
            if(temp.left != None):
                queue.append(temp.left)
            if(temp.right != None):
                queue.append(temp.right)
        return root

总结：

1. 队列和栈都可用