用递归的方式统计二叉树的结点个数同时交换二叉树每个结点的左孩子和右孩子
时间: 2023-11-15 09:05:20 浏览: 39
以下是Python代码实现:
```python
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def invertTree(root: TreeNode) -> TreeNode:
if not root:
return root
left = invertTree(root.left)
right = invertTree(root.right)
root.left = right
root.right = left
return root
def countNodes(root: TreeNode) -> int:
if not root:
return 0
left_height = get_height(root.left)
right_height = get_height(root.right)
if left_height == right_height:
return 2 ** left_height + countNodes(root.right)
else:
return 2 ** right_height + countNodes(root.left)
def get_height(root: TreeNode) -> int:
if not root:
return 0
return 1 + get_height(root.left)
```
其中,`invertTree`函数用于交换二叉树每个结点的左孩子和右孩子,`countNodes`函数用于递归统计二叉树的结点个数,`get_height`函数用于获取二叉树的高度。