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0226.翻转二叉树.md 修正几处 Python 代码不严谨说法

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problems/0226.翻转二叉树.md
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@ -81,7 +81,7 @@ if (root == NULL) return root;
3. 确定单层递归的逻辑 3. 确定单层递归的逻辑
因为是先前序遍历，所以先进行交换左右孩子节点，然后反转左子树，反转右子树。 因为是前序遍历，所以先进行交换左右孩子节点，然后反转左子树，反转右子树。
```cpp ```cpp
swap(root->left, root->right); swap(root->left, root->right);
@ -348,14 +348,13 @@ class Solution:
while stack: while stack:
node = stack.pop() node = stack.pop()
node.left, node.right = node.right, node.left node.left, node.right = node.right, node.left
if node.left:
stack.append(node.left)
if node.right: if node.right:
stack.append(node.right) stack.append(node.right)
if node.left:
stack.append(node.left)
return root return root
``` ```
递归法：中序遍历： 递归法：中序遍历：
```python ```python
# Definition for a binary tree node. # Definition for a binary tree node.
@ -374,7 +373,7 @@ class Solution:
return root return root
``` ```
迭代法：中序遍历： 迭代法，伪中序遍历（结果是对的，看起来像是中序遍历，实际上它是前序遍历，只不过把中间节点处理逻辑放到了中间。还是要用'统一写法'才是真正的中序遍历）：
```python ```python
# Definition for a binary tree node. # Definition for a binary tree node.
# class TreeNode: # class TreeNode:
@ -389,15 +388,14 @@ class Solution:
stack = [root] stack = [root]
while stack: while stack:
node = stack.pop() node = stack.pop()
if node.left: if node.right:
stack.append(node.left) stack.append(node.right)
node.left, node.right = node.right, node.left node.left, node.right = node.right, node.left # 放到中间，依然是前序遍历
if node.left: if node.right:
stack.append(node.left) stack.append(node.right)
return root return root
``` ```
递归法：后序遍历： 递归法：后序遍历：
```python ```python
# Definition for a binary tree node. # Definition for a binary tree node.
@ -416,7 +414,7 @@ class Solution:
return root return root
``` ```
迭代法：后序遍历： 迭代法，伪后序遍历（结果是对的，看起来像是后序遍历，实际上它是前序遍历，只不过把中间节点处理逻辑放到了最后。还是要用'统一写法'才是真正的后序遍历）：
```python ```python
# Definition for a binary tree node. # Definition for a binary tree node.
# class TreeNode: # class TreeNode:
@ -431,19 +429,15 @@ class Solution:
stack = [root] stack = [root]
while stack: while stack:
node = stack.pop() node = stack.pop()
if node.left:
stack.append(node.left)
if node.right: if node.right:
stack.append(node.right) stack.append(node.right)
if node.left:
stack.append(node.left)
node.left, node.right = node.right, node.left node.left, node.right = node.right, node.left
return root return root
``` ```
迭代法：广度优先遍历（层序遍历）： 迭代法：广度优先遍历（层序遍历）：
```python ```python
# Definition for a binary tree node. # Definition for a binary tree node.