按照先序遍历的顺序输出给定二叉树的叶结点
时间: 2024-05-06 20:21:40 浏览: 18
好的,理解了您的问题。以下是用 Python 语言实现的代码:
```python
class Node:
def __init__(self, val=None):
self.val = val
self.left = None
self.right = None
def preorder(node, res):
if not node:
return
if not node.left and not node.right:
res.append(node.val)
preorder(node.left, res)
preorder(node.right, res)
if __name__ == '__main__':
root = Node(1)
root.left = Node(2)
root.left.left = Node(4)
root.left.right = Node(5)
root.right = Node(3)
root.right.left = Node(6)
root.right.right = Node(7)
res = []
preorder(root, res)
for val in res:
print(val, end=' ')
```
输出结果为:4 5 6 7
以上代码通过递归实现了按照先序遍历的顺序输出给定二叉树的叶结点,输出结果即为叶结点的值。