from my_data_structure import generate_treenode, print_treenode
tc = [
generate_treenode('8,6,10,5,7,9,11'),
# 只有一个节点的二叉树
generate_treenode('1'),
# 只有左子树的二叉树
generate_treenode('1,2,null,4'),
]
def main(tree):
if tree.left == tree.right == None:
return tree
tree.left, tree.right = tree.right, tree.left
if tree.left:
main(tree.left)
if tree.right:
main(tree.right)
return tree
if __name__ == '__main__':
for t in tc:
print_treenode(main(t))
from my_data_structure import generate_treenode, print_treenode, TreeNode
tc = [
generate_treenode('5,3,6,2,4,null,7'),
generate_treenode('1,null,2'),
generate_treenode('1,null,2'),
generate_treenode('0'),
generate_treenode('5,3,6,2,4,null,7'),
generate_treenode(
'2,0,33,null,1,25,40,null,null,11,31,34,45,10,18,29,32,null,36,43,46,4,null,12,24,26,30,null,null,35,39,42,44,null,48,3,9,null,14,22,null,null,27,null,null,null,null,38,null,41,null,null,null,47,49,null,null,5,null,13,15,21,23,null,28,37,null,null,null,null,null,null,null,null,8,null,null,null,17,19,null,null,null,null,null,null,null,7,null,16,null,null,20,6'
)
]
print_treenode(tc[5])
class Solution:
def deleteNode(self, root: TreeNode, key: int) -> TreeNode:
self.head = root
def get_mid_order_next(root: TreeNode, parent: TreeNode):
"""
返回一个节点的中序遍历的下一个节点及其父节点。
注意：这里的算法只是这道题适用，因为在题里找中序遍历下一个节点时一定有右子节点，这是所有找中序遍历下一个节点的算法中的一种情况。
"""
while root.left:
return get_mid_order_next(root.left, root)
return root, parent
def delete(root, key, parent: TreeNode):
# 没找到
if not root:
return None
from my_data_structure import generate_treenode, print_treenode, TreeNode
tc = [generate_treenode('4,2,7,1,3')]
def insertIntoBST(root, val: int):
head = root
def insert(root, val):
if root.val > val:
if root.left:
insertIntoBST(root.left, val)
else:
root.left = TreeNode(val)
else:
if root.right:
insertIntoBST(root.right, val)
else:
root.right = TreeNode(val)
insert(root, val)
return head
if __name__ == '__main__':
print_treenode(insertIntoBST(tc[0], 5))
from my_data_structure import generate_treenode, print_treenode
tc = [generate_treenode('4,2,7,1,3,5,6')]
def searchBST(root, val: int):
if not root:
return None
if root.val == val:
return root
if root.val > val:
return searchBST(root.left, val)
else:
return searchBST(root.right, val)
if __name__ == '__main__':
print_treenode(searchBST(tc[0], 7))
node = node.left
return node.val
else:
if not node.parent: # 无父节点
return '最后一位'
if node.parent.left and node.parent.left.val == node.val: # 是父节点的左子节点
return node.parent.val
else: # 是父节点的右子节点
while node.parent: # 沿右子节点路线一路向上到顶
if node.parent.right.val == node.val:
node = node.parent
else:
break
if node.parent:
return node.parent.val
else:
return '最后一位'
if __name__ == '__main__':
# 打印出当前测试用例的二叉树形状
tc_tree = generate_treenode(tc, True)
print_treenode(tc_tree)
# 遍历测试每个节点的下一个节点
t, num = 1, 1 # 将num修改为二叉树中节点的个数
while t <= num:
tc_node = get_specific_treenode(tc_tree, t)
print(tc_node.val, main(tc_node), '\n')
t += 1