104. 二叉树的最大深度
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def maxDepth(self, root: Optional[TreeNode]) -> int:
if not root:
return 0
left = self.maxDepth(root.left)
right = self.maxDepth(root.right)
return max(left,right) + 1111. 二叉树的最小深度
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def minDepth(self, root: Optional[TreeNode]) -> int:
if not root:
return 0
left = self.minDepth(root.left)
right = self.minDepth(root.right)
if not root.left and root.right:
return 1+right
if not root.right and root.left:
return 1+left
return min(left,right)+1要注意：叶子节点是指没有左右子节点的节点。那么就不能无脑return min(left,right)+1 因为如果左边是空右边不是空那么会直接返回1而不是右边的长度
222. 完全二叉树的节点个数
暴力解法
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def countNodes(self, root: Optional[TreeNode]) -> int:
if not root:
return 0
res = []
queue = deque()
queue.append(root)
while queue:
size = len(queue)
for i in range(0, size):
node = queue.popleft()
res.append(node)
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return len(res)
普通二叉树写法 递归 卧槽比循环还慢！
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def countNodes(self, root: Optional[TreeNode]) -> int:
if not root:
return 0
left = self.countNodes(root.left)
right = self.countNodes(root.right)
return 1+left+right完全二叉树写法 递归 更tm慢了
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def countNodes(self, root: Optional[TreeNode]) -> int:
if not root:
return 0
count = 1
left = root.left
right = root.right
while left and right:
count +=1
left = left.left
right = right.right
if not left and not right: # 如果同时到底说明是满二叉树
return 2**count-1
return 1+self.countNodes(root.left)+self.countNodes(root.right)
