创建二叉树,输出先序中序后序
时间: 2024-01-23 22:47:13 浏览: 47
好的,下面是 Python 代码实现:
```python
# 定义二叉树节点
class Node:
def __init__(self, val):
self.val = val
self.left = None
self.right = None
# 创建二叉树
def create_tree(preorder, inorder):
if not preorder or not inorder:
return None
root_val = preorder[0]
root = Node(root_val)
index = inorder.index(root_val)
root.left = create_tree(preorder[1:index+1], inorder[:index])
root.right = create_tree(preorder[index+1:], inorder[index+1:])
return root
# 输出先序遍历
def preorder_traversal(root):
if not root:
return
print(root.val, end=' ')
preorder_traversal(root.left)
preorder_traversal(root.right)
# 输出中序遍历
def inorder_traversal(root):
if not root:
return
inorder_traversal(root.left)
print(root.val, end=' ')
inorder_traversal(root.right)
# 输出后序遍历
def postorder_traversal(root):
if not root:
return
postorder_traversal(root.left)
postorder_traversal(root.right)
print(root.val, end=' ')
# 测试
preorder = [1, 2, 4, 5, 3, 6, 7]
inorder = [4, 2, 5, 1, 6, 3, 7]
root = create_tree(preorder, inorder)
print("先序遍历:", end='')
preorder_traversal(root)
print()
print("中序遍历:", end='')
inorder_traversal(root)
print()
print("后序遍历:", end='')
postorder_traversal(root)
```
输出结果为:
```
先序遍历:1 2 4 5 3 6 7
中序遍历:4 2 5 1 6 3 7
后序遍历:4 5 2 6 7 3 1
```