def check(r, exc):
a = init_tree(r)
res = Solution().isBalanced(a)
print(res)
assert res == exc
输入: [1,null,0,0,1]
输出: [1,null,0,null,1]
解释:
只有红色节点满足条件“所有不包含 1 的子树”。
右图为返回的答案。
？ 这是中等？还是有更优的？
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/binary-tree-pruning
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
"""
from notebook.algorithm.树.utils import TreeNode
from notebook.algorithm.树.utils import init_tree
class Solution:
def pruneTree(self, root: TreeNode) -> TreeNode:
if not root:
return None
root.left = self.pruneTree(root.left)
root.right = self.pruneTree(root.right)
if not root.left and not root.right and root.val == 0:
return None
return root
a = init_tree([1, 1, 0, 1, 1, 0, 1, 0])
a = Solution().pruneTree(a)
print(a)
def check(r, exc):
a = init_tree(r)
res = Solution().countNodes(a)
print(res)
assert res == exc
from notebook.algorithm.树.utils import TreeNode
from notebook.algorithm.树.utils import init_tree
class Solution:
def searchBST(self, root: TreeNode, val: int) -> TreeNode:
if not root:
return None
if root.val == val:
return root
if root.val < val:
return self.searchBST(root.right, val)
return self.searchBST(root.left, val)
a = init_tree([4, 2, 7, 1, 3])
res = Solution().searchBST(a, 2)
print(res.val)
"""迭代"""
class Solution:
def searchBST(self, root: TreeNode, val: int) -> TreeNode:
while root is not None and root.val != val:
root = root.left if root.val > val else root.right
return root
print("迭代")
a = init_tree([4, 2, 7, 1, 3])
res = Solution().searchBST(a, 2)
class Solution:
def maxDepth(self, root: TreeNode) -> int:
if not root:
return 0
if not root.left and not root.right:
return 1
cur_depth = 0
if root.left:
cur_depth = self.maxDepth(root.left) + 1
if root.right:
right_depth = self.maxDepth(root.right) + 1
cur_depth = right_depth if right_depth > cur_depth else cur_depth
return cur_depth
a = init_tree([3, 9, 20, None, None, 15, 7])
print(Solution().maxDepth(a))
"""优雅很多"""
class Solution:
def maxDepth(self, root: TreeNode) -> int:
if not root:
return 0
return max(self.maxDepth(root.left), self.maxDepth(root.right)) + 1
a = init_tree([3, 9, 20, None, None, 15, 7])
print(Solution().maxDepth(a))
"""递归很多能通过栈实现
next_node.left = dnode.left
else:
# 否则替换节点为右子树的节点
next_node = dnode.right # 会是空
# 删除节点的父节点的子节点改为该最大节点
if pre:
if opt == 1:
pre.right = next_node
else:
pre.left = next_node
else:
root = next_node
return root
a = init_tree([5, 3, 6, 2, 4, None, 7])
res = Solution().deleteNode(a, 3)
res = Solution().deleteNode(a, 0)
a = init_tree([0])
res = Solution().deleteNode(a, 0)
a = init_tree([5, 3, 6, 2, 4, None, 7])
res = Solution().deleteNode(a, 5)
print(res.val)
# f**k you
"""
[1,0,15,null,null,4,35,3,8,25,49,2,null,5,12,22,27,47,null,null,null,null,7,11,13,19,24,26,31,40,48,6,null,9,null,null,14,17,21,23,null,null,null,30,33,39,42,null,null,null,null,null,10,null,null,16,18,20,null,null,null,28,null,32,34,36,null,41,44,null,null,null,null,null,null,null,null,null,29,null,null,null,null,null,37,null,null,43,46,null,null,null,38,null,null,45]
22
输出：
[1,0,15,null,null,4,35,3,8,25,49,2,null,5,12,21,27,47,null,null,null,null,7,11,13,19,24,26,31,40,48,6,null,9,null,null,14,17,null,23,null,null,null,30,33,39,42,null,null,null,null,null,10,null,null,16,18,null,null,28,null,32,34,36,null,41,44,null,null,null,null,null,null,null,29,null,null,null,null,null,37,null,null,43,46,null,null,null,38,null,null,45]
预期结果：
if not root.left and not root.right:
return root.val, root.val
left_min = rigth_max = root.val
if root.left:
left_min, left_max = self.isValidTree(root.left)
if left_max is None or left_max >= root.val:
return None, None
if root.right:
rigth_min, rigth_max = self.isValidTree(root.right)
# rigth_min会是子节点的数字
if rigth_min is None or rigth_min <= root.val:
return None, None
return left_min, rigth_max
a = init_tree([5, 1, 4, None, None, 3, 6])
print(Solution().isValidBST(a))
a = init_tree([2, 1, 3])
print(Solution().isValidBST(a))
a = init_tree([1, 1])
res = Solution().isValidBST(a)
print(res)
assert res == False
"""
作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/validate-binary-search-tree/solution/yan-zheng-er-cha-sou-suo-shu-by-leetcode-solution/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
"""